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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A neutron, a proton , and an electron and an alpha particle enter a region of constant magnetic field with equal velocities . The magnetic field is along the inward normal to the plane of the paper . The tracks of the particles are labelled in fig. the electron follows track .... and the alpha particle follows track..... |
| Answer» Correct Answer - A::B::C::D | |
| 102. |
A thin iron ring with mean diameter `d = 50 cm` supports a winding consisting of `N = 800` turns carrying current `I = 3.0A`. The ring has a cross-cut of width `b = 2.0 mm`. Neglecting the scattering of the magnetic flux at the gap edges, and using the plot shown in FIg. find the permeabitliy of iron under these conditions. |
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Answer» From the theorem on circulatiaon of vector `vec(H)` `H pi d + (B b)/(mu_(0)) = N I` or, `B = (mu_(0) N I)/(b) - (mu_(0) pi d)/(b) H = (1.51 - 0.987) H`, where `B` is the Tesla and `H` in `Ka//m`. Besides, `B` and `H` are interralates as in the fig of the text thus we have to solve for `B, H` graphically by simulancously drawing the two curves (the hysterists curve and the straight line, graph line, given above) and find the point of intersection. It is at `H = 0.26 kA//m, B = 1.25 T` Then, `mu = (B)/(mu_(0) H) = 4000` |
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| 103. |
An iron core shaped as a tore number of turns mean `R = 250 mm`. The supports a winding with the total number of turns `N = 1000`. The core has a cross-cut of width `b = 1.00 mm`. With a current `I = 0.85` A flowing through the winding, the magnetic induction in the gap is equal to `B = 0.75 T`. Assuming the scarttering of the magnetic flux at the grap edges to be negligible, find the permeaiility of iron under these conditions. |
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Answer» Here, `oint vec(H).d.vec(l) = N I` or, `H(2pi R) + (Bb)/(mu_(0)) = N I`, so, `H = (N I mu_(0) - Bb)/(2pi R mu_(0))` Hence, `mu = (B)/(mu_(0) H) = (2pi R B)/(mu_(0) N I - B b) = 3700` |
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| 104. |
Using the betatron condition, find the radius of a round orbit of an electron if the magnetic induction is known as a function of distance `r` from the axis of the field. Examine this problem for the specfic case `B = B_(0) - alpha r^(2)`, where `B_(0)` and `a` are positive constants. |
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Answer» The condtion, `B(r_(0)) = (1)/(2) lt B gt = (1)/(2) int_(0)^(r_(0)) B.2pi r dr//pi r_(0)^(2)` or, `B (r_(0)) = (1)/(r_(0)^(2)) int_(0)^(r_(0)) Br dr` The gives `r_(0)`. In the present case, `B_(0) - ar_(0)^(2) = (1)/(r_(0)^(2)) int_(0)^(r_(0)) (B - ar^(2)) rdr = (1)/(2) (B_(0) - (1)/(2) ar_(0)^(2))` or, `(3)/(4) ar_(0)^(2) = (1)/(2) B_(0)` or `r_(0) = sqrt((2 B_(0))/(3a))` |
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| 105. |
Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to `R = 100 mm` and the magnetic induction at its centre is equal to `B = 6.0 mu T`. |
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Answer» Magnetic moment of a current loop is given `P_(m) = niS` (where `n` is the number of turns and `S`, the current sectional area.) In our problem, `n = 1, S = pi R^(2)` and `B = (mu_(0))/(2) (i)/(R)` `P_(m) = (2 B R)/(mu_(0)) pi R^(2) = (2pi B R^(3))/(mu_(0))` |
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| 106. |
A non-conducting thin disc of radius R charged uniformly over one side with surface density `sigma`, rotates about its axis with an angular velocity `omega`. Find (a) the magnetic field induction at the centre of the disc (b) the magnetic moment of the disc. |
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Answer» (a) Let us take a ring element of radius `r` and thickness `dr`, then charge on the ring element, `d q = sigma 2 pi r dr` and current, due to this elemtn, `di = ((sigma 2pi r dr) omega)/(2pi) = sigma omega r dr` So, magentic induction at the centre, due to this element : `dB = (mu_(0))/(2) (di)/(r)` and hence , from symmetry : `B = int dB = int_(0)^(R) (mu_(0) sigma omega r dr)/(r) = (mu_(0))/(2) sigma omega R` (b) Magnetic moment of the element, considered, `dp_(m) = (dl) pi r^(2) = sigma omega dr pi r^(2) = sigma pi omega r^(3) dr` Hence, the sought magentic moment, `p_(m) = int dp_(m) = int_(0)^(R) sigma pi omega r^(3) dr = sigma omega pi (R^(4))/(4)` |
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| 107. |
A particle of mass `1xx 10^(-26) kg` and charge `+1.6xx 10^(-19) C` travelling with a velocity `1.28xx 10^6 ms^-1` in the `+x` direction enters a region in which uniform electric field E and a uniform magnetic field of induction B are present such that `E_x = E_y = 0, E_z= -102.4 kV m^-1, and B_x = B_z =0, B_y = 8X 10^-2.` The particle enters this region at time `t=0.` Determine the location (x,y,z coordinates) of the particle at `t= 5xx10^-6 s.` If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at `t= 7.45xx10^-6 s` ? |
| Answer» Correct Answer - A | |
| 108. |
An electron gun `G` emits electons of energy `2 keV` travelling in the positive x-direction. The electons are required to hit the spot `S` where `GS=0.1 m`, and the line `GS` makes an angle of `60^@` with the x-axis as shown in figure. A uniform magnetic field `B` parallel to `GS` exists in the region outside the electron gun. find the minimum value of `B` needed to make the electrons hit `S`. |
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Answer» Correct Answer - A::D `t=1/(V cos 60), T=(2pim)/(qB)` `nT=t` `(n2pimV)/(qB)=0.2` `(pinsqrt(2mE))/(qB)=0.1` `B=(pinsqrt(2xx9xx10^(-31)xx2xx10^(3)xx1.6xx10^(-19)))/(exx0.1) rArrB=(pin)/0.1sqrt((36xx10^(-28))/(1.6xx10^(-19)))=(pin)/0.1sqrt((9xx10^(-6))/4)` `B=15pinxx10^(-4)T rArr B_(min)=15pixx10^(-4)T` |
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| 109. |
Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference `V` along `X`-axis.These electrons emerges from a narrow hole into a unifrom magnetic field `B` directed along the axis.However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure.These paraxial electrons meet for second time on the `X`-axis at a distance `sqrt((Npi^(2)mV)/(eB^(2)))`.Then find value of `N`.(Neglect interaction between electrons) |
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Answer» Correct Answer - B::C Pitch `=(2pimcostheta)/(qB) eV=1/2mv^(2)` `=(2pisqrt(2meV))/(eB) (costheta~~1) rArr "pitch" =sqrt((8pi^(2)mV)/(eB^(2)))` `:.` Distance from point of divergence `=sqrt((32pi^(2)mV)/(eB^(2)))` |
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| 110. |
A spherical shell of radius `R_(1)` with uniform charge q is expanded to a radius `R_(2)`. Find the work performed by the electric forces in this process. |
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Answer» As the field is conservative total work done by field force, `A_(fd) = U_(i) - U_(f) = (1)/(2) q (varphi_(1) - varphi_(2))` `= (1)/(2) (q^(2))/(4pi epsilon_(0)) [(1)/(R_(1)) - (1)/(R_(2))] = (q^(2))/(8pi epsilon_(0)) [(1)/(R_(1)) - (1)/(R_(2))]` |
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| 111. |
A positive point charge `50 mu C` is located in the plane `xy` at the point with radius vector `r_(0) = 2i + 3j`, where `I` and j are the unit vectors of the `x` and `y` axes. Find the vector of the electric field strength `E` and its magnitude ath the point with radius vector `r = 8i - 5j`. Here `r_(0)` and `r` are expressed in metres. |
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Answer» Sought field strength `vec(E) = (1)/(4pi epsilon_(0)) (q)/(|vec(r ) - vec(r_(0))|^(2))` `= 4.5 kV//m` on putting the values. |
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| 112. |
In a betatron the magnetic flux across an equilibrium orbit of radius `r = 25 cm` grows during the acceleration time at paractically constant rate `Phi = 5.0 Wb//s`. In the process, the electrons acquire an energy `W = 25 MeV`. Find the number of revolutions made by the electron during the acedeleration time and the corresponding distance covered by it. |
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Answer» The increment is energy per revoltion is `e Phi`, so the number of revoltutions is, `N = (W)/(e Phi)` The distance traversed is, `s = 2pi rN` |
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| 113. |
A thin wire ring of radius `a` carrying a charge `q` approcahes the observation point `P` so that its centre moves rectilinearly with a constant velocity `v`. The plane of the ring remians perpendicular to the motion direction. At what distance `x_(m)` from the point `P` will the ring be located at the moment when the displacement current density at the point `P` becomes maximum? What is the magnitude of this maximum density ? |
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Answer» We have, `E_(P) = (q x)/(4pi epsilon_(0) (a^(2) + x^(2))^(3//2))` then `j_(4) = (del D)/(del t) = epsilon_(0) (del E)/(del t) = (qv)/(4pi (a^(2) + x^(2))^(5//2)) (a^(2) - 2x^(2))` This is maximum, when `x = x_(m) = 0`, and minimum at some other value. The maximum displacement current density is `(j_(d))_(max) = (qv)/(4pi a^(3))` To check this we calculate `(del j_(d))/(del x)`, `(del j_(4))/(del x) = (q v)/(4pi) [(-4x (a^(2) + x^(2)) -5x(a^(2) - 2x^(2))]` This vanishes for `x = 0` and for `x = sqrt((3)/(2)) a`. The latter is easily shown to be a smaller local minimum (negative maximum). |
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| 114. |
A 50 turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) in a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil? |
| Answer» Correct Answer - A::B::D | |
| 115. |
A ring of mass `m` and radius `r` is rotated in uniform magnetic field `B` which is perpendicular to the plane of the loop with constant angular velocity `omega_(0)`.Find the net ampere force on the ring and the tension developed in the ring if there is a current `i` in the ring.Current and rotation both are clockwise. |
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Answer» Correct Answer - `0,r/(2pi)(momega_(0)^(2)+2piiB)` Net ampere force acting on a closed loop in uniform magnetic field is zero. `2Tsin("d" theta)/2=(dm.r.omega^(2))+r."d" delta.iB` `T."d" delta=m/(2pir).rd theta.r.omega^(2)+r."d" delta i.B` `T=r/(2pi)(momega_(1)^(2)+2pi i.B)`. ` ` |
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| 116. |
A straight round copper conductor of radius `R = 5.0 mm` carries a current `I = 50 A` Find the potentail difference between the axis of the conductor and its surface. The concentration of the conduction electrons in copper is equal to `n = 0.9.10^(33) cm^(-3)`. |
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Answer» The electrons in the conductor are drifting with a speed of, `v_(d) = (J)/("ne") = (I)/(pi R^(2) "ne")`, where `e` = magnitude of the charge on the electron, `n` = concentration of the conductin electrons. The magentic field inside the conductor due to this current is given by, `B_(varphi) (2pi r) = pi r^(2) (I)/(pi R^(2)) mu_(0)` or, `B_(varphi) = (mu_(0))/(2pi) (I r)/(R^(2))` A radial electric field `vB_(varphi)` must come into being in equilibrium. Its `P.D.` is, `Delta varphi = int_(0)^(R) (I)/(pi R^(2) n e) (mu_(0))/(2pi) (Ir)/(R^(2)) dr = (I)/(pi R^(2)n e) ((mu_(0))/(4pi) I) = (mu_(0) I^(2))/(4pi R^(2) n e)` |
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| 117. |
Find the electric field strength vector if the potentail of this field has the form `varphi = ar`, where is a constatn vector, and `r` is the radius vector of point of the field. |
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Answer» In accordance with the problem `varphi = vec(a). vec(r )`. Thus from the equation : `vec(E) = vec(grad) varphi` `vec(E) = -[(del)/(del x) (a_(x) x) vec(i) + (del)/(del_(y)) vec(j) + (del)/(del_(z)) (a_(z) z) vec(k)] = -[a_(x) vec(i) + a_(y) vec(j) + a_(z) vec(k)] = -vec(a)` |
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| 118. |
The potential of a certain electrostaitc field has the form `varphi = a (x^(2) + y^(2)) + bz^(2)`, where `a` and `b` are constants. Find the magnitude and direction fo the electric field strength vector. What shape have the equipotentail surfaces in the fololowing cases: (a) `a gt 0, b gt 0`, (b) `a gt 0, b lt 0` ? |
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Answer» Given, `varphi = a (x^(2) + y^(2)) + bz^(2)` So, `vec(E) = -vec(grad) varphi = -[2 ax vec(i) + 2ay vec(j) + 2 bz vec(k)]` Hence `|vec(E)| = 2 sqrt(a^(2) (x^(2) + y^(2)) + b^(2) z^(2))` Shape of the equipotential surface : Put `vec(rho) = x vec(i) + y vec(j)` or `rho^(2) = x^(2) + y^(2)` Then the quipotential surface has the equaction `a rho^(2) + b z^(2)` = constant `= varphi` If `a gt 0, b gt 0` then `varphi gt 0` and teh equaction of the equipotential surface is `(rho)/(varphi//a) + (z^(2))/(varphi//b) = 1` which is an ellipose in `rho, z` coordinates. In three dimensions the surface is an elipsoid of revolution with semi-axis `sqrt(varphi//a), sqrt(varphi//a), sqrt(varphi//b)`. If `agt0, blt0` then `varphi` can be `ge 0`. If `varphi gt 0` then the equaction is `(rho^(2))/(varphi//a) - (z^(2))/(varphi//|b|) = 1` This si a single cavity hyperboloid of revolution about `z` axis. If `varphi = 0` then `a rho^(2) = |b| z^(2) = 0` or `z = +- sqrt((a)/(|b|)) rho` is the equaction of a right circular cone. If `varphi lt 0` then the equaction can be written as `|b| x^(2) - a rho^(2) = |varphi|` or `(x^(2))/(|varphi|//|b|) = (rho^(2))/(|varphi|//a) = 1` This is a two cavity hyperbotold of revolution about z-axis. |
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| 119. |
Find the potential `varphi (x,y)` of an electrostatic field `E = 2axyi + a (x^(4) - y^(2)) j`, where `a` is a constant, `I` and `J` are the unit vectors of the `x` and `y` axes. |
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Answer» `-d varphi = vec(E) . vec(dr) = [2 a xy vec(i) + 2(x^(2) - y^(2)) vec(i)] . [dx vec(i) + dy vec(i)]` or, `d varphi = 2a xy dx + a(x^(2) - y^(2)) dy = ad (x^(2) y) - ay^(2) dy` On intergating, we get `varphi = ay ((y^(2))/(3) - x^(2)) + C` |
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| 120. |
(a) Soft iron is a conductor of electricity. (b)It is a magnetic material. ( c)It is an alloy of iron. (d)It is used for making permanent magnets.State whether:A. `a` and `c` are trueB. `a` and `b` are trueC. `c` and `d` are trueD. `b` and `d` are true |
| Answer» Correct Answer - B | |
| 121. |
Soft iron is used in many electrical machines for:A. low hysteresis loss and low permeabilityB. low hysteresis loss and high permeabilityC. high hysteresis loss and low permeabilityD. high hysteresis loss and high permeability |
| Answer» Correct Answer - B | |
| 122. |
A capacitor of capacitance `C_(1) = 1.0 muF` carrying initially a voltage `V = 300 V` is connected in parallel with an uncharged capacitor of capacitance `C_(2) = 2.0 muF`. Find the increment of the electric energy of this system by the moment equilibrium is reached. Explain the result obtained. |
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Answer» Charge contained in the capacitor fo capacitance n`C_(1)` is `q = C_(1) varphi` and the energy, stored in it : `U_(i) = (q^(2))/(2 C_(1)) = (1)/(2) C_(1) varphi^(2)` Now, when the capacitor are connected in parallel, equivalent capacitance of the system, `C = C_(1) + C_(2)` and hence, energy stored in the system : `U_(f) = (C_(1)^(2) varphi^(2))/(2 (C_(1) + C_(2)))`, as charge remains conserved during the process. So, increment in the energy, `Delta U = (C_(1)^(2) varphi^(2))/(2) ((1)/(C_(1) + C_(2)) - (1)/(C_(1))) = (-C_(2) C_(1) varphi^(2))/(2 (C_(1) + C_(2))) = 0.03 mJ` |
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| 123. |
A capaitor of capacitance `C_(1) = 1.0 muF` withstands teh maximum voltage `V_(1) = 6.0 kV` while a capacitor of capacitance `C_(s) = 2.0 muF`, the maximum voltage `V_(s) = 4.0 kV`. What voltage will the system of these two capacitors withsatand if they are connected in sereis ? |
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Answer» Amount of charge, that the capacitor of capacitance`C_(1)` can withstand, `q_(1) = C_(1) V_(1)` and similarly the charge, that the capacitor of capacitance `C_(2)` can withstand , `q_(2) = C_(2) V_(2)`. But in series combination, charge on both the capacitors will be same, so `q_(max)`, that the conservation can withstand `= C_(1) V_(1)`, as `C_(1) V_(1) lt C_(2) V_(2)`, from the numberical data, given. Now, net capacitance of the system, `C_(0) = (C_(1) C_(2))/(C_(1) + C_(2))` and hence, `V_(max) = (q_(max))/(C_(0)) = (C_(1) V_(1))/(C_(1) C_(2)//C_(1) + C_(2)) = V_(1) (1 + (C_(1))/(C_(2))) = 9 kV` |
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| 124. |
A capacitor with capacitance `C = 400 pF` is connected via a resistance `R = 650 Omega` to a source of voltage `V_(0)`. How soon will the voltage developed across the capacitor reach a value `V = 0.90 V_(0)` ? |
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Answer» The formula is, `q = C V_(0) (1 - e^(-1//RC))` or, `V = (q)/(C) = V_(0) (1 - e^(-1//RC))` or , `(V)/(V_(0)) = 1 - e^(-1//RC)` or, `e^(-1//RC) = 1 - (V)/(V_(0)) = (V_(0) - V)/(V_(0))` Hence, `t = RC` In `(V_(0))/(V_(0) - V) = R C` In 10, if `V = 0.9 V_(0)` Thus `t = 0.6 muS`. |
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| 125. |
A point charge `q` moves with a non-relatives velocity v = const. Find the displacement current density `dotd` at a point location at a distance `r` from the charges on a straight line (a) coinciding with the charge path, (b) perpendicualr to the path and passing through the charge. |
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Answer» In the non-relativistic limit. `vec(E) = (q)/(4pi epsilon_(0) r^(3)) vec(r)` (a) On a striaght line coinciding with the charge path, `vec(j_(4)) = epsilon_(0) (del vec(E))/(del t) = (q)/(4pi) [(-vec(V))/(r^(3)) - (3 vec(r) r)/(r^(4))], ("using", (d vec(r))/(dt) = -vec(v).)` But in this case, `dotr = -v` and `v (vec(r))/(r) = vec(v)`, so, `j_(d) = (2q vec(v))/(4pi r^(3))` (b) In this case, `dotr = 0`, as, `vec(r) _|_ vec(v)`. Thus, `j_(4) = (q vec(v))/(4pi r^(3))` |
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| 126. |
The air between two parallel plates separated by a distance `d = 20 mm` is inoized by X-ray radisation. Each plate has an area `S = 500 cm^(2)`. Find the concentration of positive ions if at a voltage `V = 100 V` a current `I = 30 muA` flows between the paltes, which is well below the saturation current. The air ion mobilities are `u_(0)^(+) = 1.37 cm^(2)//(V .s)` and `u_(0) = 1.91 xm^(2)//(V . s)` |
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Answer» `E = (V)/(d)` So by the definition of the mobility `v^(2) = u_(0)^(+) (V)/(d), v^(-) = u_(0)^(-) - (V)/(d)` and `j = (n_(+) u_(0)^(+) + n_(-) u_(0)^(-)) (eV)/(d)` (The negative ions move towards the anode and the positive ion towards the cathode and the total current is the sum of the currents due to them.) On the other hand, in equilibrium `n_(+) = n_(-)` So, `n_(+) = n_(-) = (I)/(S)//(u_(0)^(+) + u_(0)^(-)) (eV)/(d)` `= (Id)/(e V S (u_(0)^(+) + u_(0)^(-))) = 2.3xx10^(8) cm^(-3)` |
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| 127. |
A gas is ionized in the immeiate vicinity of the surface of plane electrode `I` (Fig) separated from electrodes 2 by a distanae `l`. An alternating voltage varying with time `t` as `V = V_(0) sin omega t` is applie dto the electrodes. On decreasing the frequency `omega` it was observed that the galvonometer `G` indicates a current only at `omega lt omega_(0)`, where `omega_(0)` is a certain cut-off frequency. FInd the mobility of ions reaching electrode 2 under these conditions. |
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Answer» Velocity = mobility `xx` field or, `v = u (V_(0))/(l) sin omega t`, which is positive for `0 le omega t le pi` So, maximum displacement in one direction is `x_(max) = int_(0)^(pi) u (V_(0))/(l) sin omega t dt = (2 u V_(0))/(l omega)` At `omega = omega_(0), x_(max) = l`, so `(2u V_(0))/(l omega) = l` Thus `u = (omega l^(2))/(2 V_(0))` |
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| 128. |
The space between two conducting concentric spheres of radii `a` and `b (a lt b)` is filled up with homongeneous poorly conducting medium. The capacitance of such a system equals `C`. Find the resistivity of the medium if the potential difference between the spheres, when they are disconnected from an external voltage , decreases `eta`-fold during the time interval `Delta t`. |
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Answer» In our system, resistance of the medium `R = (rho)/(4pi) [(1)/(a) - (1)/(b)]` where `rho` is the resistivity of the medium The current `i = (varphi)/(R) = (varphi)/((rho)/(4pi) [(1)/(a) - (1)/(b)])` Also, `i = (-dq)/(dt) = - (d(C varphi))/(dt) = -C (d varphi)/(dt)`, as capacitance is constant. ....(2) So, equating (1) and (2) we get, `(varphi)/((rho)/(4pi) [(1)/(a)-(1)/(b)]) = -C (d varphi)/(dt)` or, `-int (d varphi)/(varphi) = (Delta t)/((C rho)/(4pi) [(1)/(a) - (1)/(b)])` or, In `eta = (Delta t 4pi ab)/(C rho (b-a))` Hence, resistivity of the medium, `rho = (4pi Delta t ab)/(C (b-a) In eta)` |
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| 129. |
A system consists of two concentric spheres, with the inside sphere of radius a carrying a positive charge `q_(1)`. What charge `q_(2)` has to be diposited on the outsie sphere of radius `b` to reduce the potentail `varphi` depend in this case on a distance `r` from the centre of the system ? Draw teh appoximate plot of this dependence. |
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Answer» Potentail at the inside sphere, `varphi_(a) = (q_(1))/(4pi epsilon_(0) a) + (q_(2))/(4pi epsilon_(0) b)` Obviously `varphi_(a) = 0` for `q_(2) = - (b)/(a) q_(1)` ...(1) When `r ge b`, `varphi = (q_(1))/(4pi epsilon_(0) r) + (q_(2))/(4pi epsilon_(0) r) = (q_(1))/(4pi epsilon_(0)) (1 - (b)/(a))//r`, using Eq. (1). And when `r le b` `varphi_(r) = (q_(1))/(4pi epsilon_(0) r) + (q_(2))/(4pi epsilon_(0) b) = (q_(1))/(4pi epsilon_(0)) ((1)/(r ) - (1)/(a))` |
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| 130. |
A dipole with an electric moment p is located at a distance r from a along thread charge uniformly with a linear density `lambda`. Find the force F acting on the dipole if the vector p is oriented (a) along the thread (b) along the radius vector r (c ) at the right angles to the thread and the radius vector r. |
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Answer» Here, `E_(r) = (lambda)/(2pi epsilon_(0) r), E_(0) = E_(varphi) = 0` and `vec(F) = p (del vec(E))/(del l)` (a) `vec(p)` along the thread. `vec(E)` does not change as the point of observation is moved along the thread. (b) `vec(p)` along `vec(r)`, `vec(F) = F_(r) vec(e_(r)) = (lambda p)/(2pi epsilon_(0) r^(2)) vec(e_(r)) = (lambda vec(p))/(2pi epsilon_(0) r^(2))` (on using `(del)/(del r) vec(e_(r)) = 0`) (c) `vec(p)` along `vec(e_(theta))` `vec(F) = p (del)/(r del theta) (lambda)/(2pi epsilon_(0) r) e_(r)` `= (p lambda)/(2pi epsilon_(0) r^(2)) (del vec(e_(r)))/(del theta) = (p lambda)/(2po epsilon_(0)r^(2)) vec(e_(theta)) = (vec(p) lambda)/(2pi epsilon_(0) r^(2))` |
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| 131. |
An electric field of strength `E = 1.0 kV//cm` produces plarization in water equivalent to the correct orientation of only one out of `N` . The electric moment of a water molecule equals `p = 0.62.10^(-29)` C.m. |
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Answer» The total polarization is `P = (e - 1) epsilon_(0) E`. This must equals `(n_(0) p)/(N)` where `n_(0)` is the concertaition of water molecules. Thus `N = (n_(0) P)/((e - 1) epsilon_(0) E) = 2.93xx10^(3)` on putting the values |
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| 132. |
The gap between the plates of a parallel-plate capacitor is filled up with an inhomongeous poorly conducting medium whose conductivity varies lineraly in the direaction perpendicular to the plates form the `sigma_(1) = 1.0 pS//m to sigma_(2) = 2.0 pS//m`. Each plate has an area `S = 230 cm^(2)`, and the separation between the plates is `d = 2.0 mm` . Find teh current flowing thorugh the capacitor due to a voltage `V = 300 V`. |
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Answer» The resistance of a layer of the medium, of thickness `dx` and at a distance `x` from the first plate of the capacitor is given by, `dR = (1)/(sigma (x)) (dx)/(S)` ....(1) Now, since `sigma` varies linerly with the distance from the plane. It may be represented as, `sigma = sigma_(1) + ((sigma_(2) - sigma_(1))/(d)) x`, at a distance `x` from any one of plate. From Eq. (1) `dR = (1)/(sigma_(1) + ((sigma_(2) - sigma_(1))/(d))x) (dx)/(S)` or, `R = (1)/(S) int_(0)^(d) (dx)/(sigma_(1) + ((sigma_(2) - sigma_(1))/(d))x) = (d)/(S(sigma_(2) - sigma_(1))) In (sigma_(2))/(sigma_(1))` Hence, `i = (V)/(R) = (S V (sigma_(2) - sigma_(1)))/(d In (sigma_(2))/(sigma_(1))) = 5 nA` |
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| 133. |
The gap between the plates of a parallel-plate capacitor is filled with glass of resistivity `rho = 100 G Omega m`, The capacitance of the capacitors equals `C = 4.0 nF`. Find the leakeage current of the capacitor when a voltage `V = 2.0 kV` is applied to it. |
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Answer» Let us mentally impart the charges `+q` and `-q` to the plates of the capacitor. Then capacitance of the network , `C = (q)/(varphi) = (epsilon epsilon_(0) int E_(n) dS)/(varphi)` ....(1) Now, electric current, `i = int vec(j) d vec(S) = int sigma E_(n) as vec(j) uarr uarr vec(E)`. ....(2) Hence, using (1) IN (2), we get, `i = (C varphi)/(epsilon epsilon_(0)) = 1.5 mu A` |
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| 134. |
A non-polar molecule with polarzabillity `beta` is located at a great distance `l` from a polar molecule with electric moment `p`. Find the magnitude of the interaction force between the molecules if the vector `p` si oriented along a straight line passing through both molecules. |
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Answer» From the genral formula `vec(E) = (1)/(4pi epsilon_(0)) (3 vec(p). vec(r ). vec(r ) - vec(p) r^(2))/(r^(5))` `vec(E) = (1)/(4pi epsilon_(0)) (2vec(p))/(l^(3))`, where `r = l`, and `vec(r ) uarr uarr vec(p)` This will cause the induction of a dipole moment. `vec(p_("ind")) = beta (1)/(4pi epsilon_(0)) (2 vec(p))/(l^(3)) xx epsilon_(0)` Thus the force, `vec(F) = (beta)/(4pi ) (2p)/(l^(3)) (del)/(del l) (1)/(4pi epsilon_(0)) (2p)/(l^(3)) = (3 beta p^(2))/(4pi^(2) epsilon_(0) l^(7))` |
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| 135. |
Initially the space between the plates of the capacitor is filled with air, and the field strength in the gap is equal to `E_(0)`. Then half the gap is filled with uniform isotropic dielectric with permittivity `epsilon` as shown in Fig. Find the moduli of the introduction of the dielectric (a) deos not change the voltage across the plates, (b) leaves the charges at the plates constant. |
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Answer» (a) We have `D_(1) = D_(2)`, or epsilon `E_(2) = E_(1)` Also, `E_(1) (d)/(2) + E_(2) (d)/(2) = E_(0) d` or, `E_(1) + E_(2) = 2 E_(0)` Hence, `E_(2) = (2 E_(0))/(epsilon + 1)` and `E_(1) = (2epsilon E_(0))/(epsilon + 1)` and `D_(1) = D_(2) = (2 epsilon epsilon_(0) E_(0))/(epsilon + 1)` (b) `D_(1) = D_(2)`, or `epsilon E_(2) = E_(1) = (sigma)/(epsilon_(0)) = E_(0)` Thus, `E_(1) = E_(0), E_(2) = (E_(0))/(epsilon)` and `D_(1) = D_(2) = epsilon_(0) E_(0)` |
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| 136. |
Under certain condinates the polarrization of an infinite uncharged dielectric plate , takes the form `P = P_(0)` is a vector perpendicular to the plate, `x` is the distance from the of the electric field inside the plate and the potentail difference between its surface. |
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Answer» Since there are no free extraneous chagres anywhere `di v vec(D) = (del D)/(del x) = 0` or, `D_(x)` = constant But `D_(x) = 0` at `oo`, so `D_(x) = 0`, every where. Thus, `vec(E) = - (vec(P_(0)))/(epsilon_(0)) (1 - (x^(2))/(d^(2)))` or `E_(x) = (P_(0))/(epsilon_(0)) (1 - (x^(2))/(d^(2)))` So, `varphi = (P_(0)x)/(epsilon_(0)) - (P_(0) x^(3))/(3 epsilon_(0) d^(2)) +` constant Hence, `varphi(+d) - varphi(-d) = (2P_(0))/(epsilon_(0)) - (2 P_(0) d^(3))/(3 d^(2) epsilon_(0)) = (4 P_(0) d)/(3 epsilon_(0))` |
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| 137. |
In a region magnetic field along `x` axis changes with time according to the given graph. If time period, pitch and radius of helix path are `T_(0),P_(0)` and `R` respectively then which of the following is incorrect if the paticle is projected at an angle `theta_(0)` with the positive `x`-axis in `x-y` plane from origin: A. At `t=T_(0)/2`,co-ordinates of charge are `(P_(0)/2,0,-2R_(0))`.B. At `t=3T_(0)/2`,co-ordinates of charge are `(P_(0)/2,0,-2R_(0))`.C. Two extremes from `x`-axis are at a distance `2R_(0)` from each other.D. Two extremes from `x`-axis are at a distance `4R_(0)` from each other. |
| Answer» Correct Answer - C | |
| 138. |
A glass plate totally fills up the gap between the electrodes of a prallel-plate capacitor whose capacitance in the absence of that glass plate is equal to `C = 20 nF`. The capacitor is connected that glass voltage source `V = 100 V`. The plate is slowly, and without friction, extracted from the gap. Find teh capacitor energy increment and the mechanical work perfomed in the process of plate extraction. |
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Answer» Initailly, capacitance of the system `= C epsilon` So, initial energy of the system : `U_(i) = (1)/(2) (C epsilon) V^(2)` and finally, energy of the capacitor : `U_(f) = (1)/(2) C V^(2)` Hence capacitance energy increment, `Delta U = (1)/(2) CV^(2) - (1)/(2) (C epsilon) V^(2) = - (1)/(2) CV^(2) (epsilon - 1) = 0.5 mJ` From enerfgy conservation `Delta U = A_("cell") + A_("agent")` (as there is no heat liberation) But `A_(cell) = (C_(f) - C_(i)) V^(2) = (C - C epsilon)V^(2)` Hence `A_("agent") = Delta U - A_(cell)` `= (1)/(2) C (1 - epsilon) V^(2) = 0.5 m J` |
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| 139. |
A rigid wire consists of a semicircular portion of radius R and two straight sections the wire is partially immersed in a perpendicualr magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current `i`. |
| Answer» Correct Answer - A::B | |
| 140. |
A straight copper wire of length `l = 1000 m` and cross sectional area `S = 1.0 mm^(2)` carries a current `I = 4.5 A`. Assuming that one free electron corresponds to each copper atom, find, (a) the time it takes an electron to displace from one end of the wire to the other, (b) the sum of electric froces acting on all free electrons in the given wire. |
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Answer» Let, `n` be the volume density of electrons, then from `I = rho S_(x) V_(d)` `I = n e S_(x)| lt vec(v) gt | = n e S_(x) (l)/(t)` So, `t = (n e S_(x) l)/(I) = 3 mu s`. (b) Sum of electrons froces `= | (nv) e vec(E)| = |n S l e rho vec(j)|`. where `rho` is resistivity of the material. `= n S l epsilon rho (I)/(S) = n e l rho I = 1.0 muN` |
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| 141. |
A wire loop enclosing as semicircle of radius `R` is located on the boudary of uniform magnetic field `B`. At the moment `t=0`, the loop is set into rotation with a costant angular acceleration `alpha` about an axis `O` coinciding with a line of vector B on the boundary. Find the emf induced in the loop as a function of time. Draw the approximate plot of this function.The arrow in the figure shows the emf direction taken to be positive. |
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Answer» Flux at any moment of time, `|Phi_(t)| = |vec(B). D vec(S)| = B ((1)/(2) R^(2) varphi)` where `varphi` is the sector angle, enclosed by the field Now, magnitude of induced `e.m.f` is given by, `xi_("in") = |(d Phi_(t))/(dt)| = |(B R^(2))/(2) (d varphi)/(dt)| = (BR^(2))/(2) omega`, where `omega` is the angular velocity of the disc. But as it series rotating from rest at `t = 0` with an angular accelearation `beta` its in angular velocity `omega (t) = beta t`. So, `xi_(i n) = (B R^(2))/(2) beta t`. According to Lenz law the first half cycle current in the loop is in anticlockwise sense, and in subsequent half cycle it is in clockwise sense. Thus is genral, `xi_(i n) = (-1)^(n) (B R^(2))/(2) beta t`, where `n` in number of half revolutions. The plot `xi_(i n) (t)`, where `t_(n) = sqrt(2 pi//beta)` is shown in the answer sheet. |
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| 142. |
For the circuit shown in figure the direction and magnitude of the force on `PQR` is : A. No resultant force act on the loopB. `ILB` out of the pageC. `1/2ILB` into the pageD. `ILB` into the page |
| Answer» Correct Answer - A | |
| 143. |
if a charged particle projected in a gravity free room it does not deflect.A. electric field and magnetic field must be zeroB. both electric field and magnetic field may be present.C. electric field will be zero and magnetic field may be zeroD. electric field may be zero and magnetic field must be zero. |
| Answer» Correct Answer - B | |
| 144. |
A negative charged particle falling freely under gravity enters a region having uniform horizontal magnetic field pointing towards north.The particle will be deflected towardsA. EastB. WestC. NorthD. South |
| Answer» Correct Answer - B | |
| 145. |
A particle with specific charge `q//m` moves rectilinerarly due to an electric field `E = E_(0) - ax`, where `a` is a positive constant, `x` is the distance from the point where the particle was initially at rest. Find: (a) the distance covered by the particle till the moment it came to a standstill, (b) the acceleration of the particle at that moment. |
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Answer» The equaction of motion is, `(dv)/(dt) = v (dv)/(dx) = (q)/(m) (E_(0) - ax)` Intergrating `(1)/(2) v^(2) - (q)/(m) (E_(0) x - (1)/(2) ax^(2))` = constant. But initallty `v = 0` when `x = 0`, so "constant" = 0 Thus, `v^(2) = (2q)/(m) (E_(0) x - (1)/(2) ax^(2))` Thus, `v = 0` again for `x = x_(m) = (2 E_(0))/(a)` The corresponding accleration is, `((dv)/(dt))_(x_(m)) = (q)/(m) (E_(0) - 2 E_(0)) = - (q E_(0))/(m)` |
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| 146. |
A long straight wire carrying a current `I` and a `II`- shaped conductor with siding connector are located in the same plane as shown in Fig. The connector of length `l` and resistance `R` sides to the right with a current induced in the loop as a function of separation `r` between the connector and the stragiht wire. The resistance of the `II`-shaped conductor and the self-induced of the loop are assumed to be negligible. |
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Answer» Field, due to the current carrying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by, `B = (mu_(0))/(2pi) (i)/(r)` where `r` is the perpendicular distance from the wire. As `B` is same length of the rod thus motinal e.m.f `xi_(i n) = |-int_(1)^(2) (vec(v) xx vec(B)) . d vec(l)| = vB l` and it is directed in the sense of `(vec(v) xx vec(B))` So, current (induced) in teh loop, `i_(i n) = (xi_(i n))/(R) = (1)/(2) (mu_(0) I vi)/(pi R r)` |
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| 147. |
Determine the accelration of a relativistic electron moving along a unifrom electric field of strength `E` at the moment when its kinetic energy becomes equal to `T`. |
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Answer» As before, `T = e E x` Now in linear motion, `(d)/(dt) (m_(0) v_(x))/(sqrt(1 - v^(2)//c^(2))) = (m_(0) w)/(sqrt(1 - v^(2)//c^(2))) + (m_(0) w)/((1 - v^(2)//c^(2))^(3//2)) = (v)/(c^(2)) w` `= (m_(0))/((1 - v^(2)//c^(2))^(3//2))w = ((T + m_(0) c^(2))^(3))/(m_(0)^(2) c^(6)) w = e E`, So, `w = (e E m_(0)^(2) c^(6))/((T + m_(0) c^(2))^(3)) = (e E)/(m_(0)) (1 + (T)/(m_(0) c^(2)))^(-3)` |
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| 148. |
Two coaxial solenoids of different radius carry current I in the same direction. `vec(F_1)` be the magnetic force on the inner solenoid due to the outer one and `vec(F_2)` be the magnetic force on the outer solenide due to the inner one. ThenA. `barF_(1)=barF_(2)=0`B. `barF_(1)` is radially inwards and `barF_(2)` is radially outwardsC. `barF_(1)` is radially inwards and `barF_(2)=0`D. `barF_(1)` is radially outwards are `barF_(2)=0` |
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Answer» Correct Answer - 1 `vecF_(1)=vecF_(2)=0` Because net resultant will be zero.and equal because of action and reaction pair |
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| 149. |
A conductor lies along the ` z`-axis at ` -1.5 le z lt 1.5 m` and carries a fixed current of `10.0 A` in `- hat(a)_(z)` direction ( see figure). For a field `vec(B) = 3.0 xx 10^(-4) e^(-0.2x) hat(a)_(y) T`, find the power required to move the conductor at constant speed to `x = 2.0 m , y = 0 m ` in `5xx 10^(-3)s` . Assume parallel motion along the ` x-axis`. A. `1.57 W`B. `2.97 W`C. `14.85 W`D. `29.7 W` |
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Answer» Correct Answer - 2 `F_(ext)=B(x)IL` `P=1/tunderset(0)overset(2)intF_(ext)dx=1/tunderset(0)overset(2)intB(x)ILdx=1/(5xx10^(-3))underset(0)overset(2)int3xx10^(-4)e^(-0.2x)xx10xx3dx` `=9[1-1/e^(0.4)]=2.96` |
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| 150. |
A point charge `q` is located at the centre `O` of a spherical uncharged coducting layer provided with small orifice. The inside and outside radii of the layer are equal to a and `b` respectively. The amount of work that has to be performed to slowly transfer the charge `q` from teh point `O` through the orifice and into infinity is |
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Answer» Initally, there will be induced charges of magnitude `-q` and `+q` on the inner and outer surface of the spherical layer respectively. Hence, the total electrical energy of the system is the sum of self energies of spherical shells, having radii `a` and `b`, and their mutal energies including the point charge `q`. `U_(i) = (1)/(2) (q^(2))/(4pi epsilon_(0) b) + (1)/(2) ((-q)^(2))/(4pi epsilon_(0) a) + (-q q)/(4pi epsilon_(0) a) + (qq)/(4pi epsilon_(0) b) + (-qq)/(4pi epsilon_(0) b)` or `U_(i) = (q^(2))/(8pi epsilon_(0)) [(1)/(b) - (1)/(a)]` Finally, charge `q` is at infinity hence, `U_(f) = 0` Now, work done by the great = increment in the energy `= U_(f) - U_(i) = (q^(2))/(8pi epsilon_(0)) [(1)/(a) - (1)/(b)]` |
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