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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.
| 51. |
A cylindrical capacitor is filled with two cyclindrical layers of dielectric with permittivity `espilon_(1)` and `espilon_(2)`. The inside radii of the layers are equal to `R_(1)` and `R_(2) gt R_(1)`. The maximum permissible values of electric field strength reaching teh breakdown value for both dielectrics simulttaneously ? |
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Answer» Let `lambda` be the linear charge density then, `E_(1m) = (lambda)/(2pi epsilon_(0) R_(1) epsilon_(1))` ….(1) and `E_(2m) = (lambda)/(2pi epsilon_(0) R_(2) epsilon_(2))` ….(2) The breakdown in either case will occur at the smaller value of `r` for a simulataneous breakdown of both dielectrics. From (1) and (2) ltbr. `E_(1m) R_(1) epsilon_(1) = E_(2m) R_(2) epsilon_(2)`, which is the sougth relationship. |
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| 52. |
A cylindrical layer of dielectric with permittivity `epsilon` is inserted into a cylindrical capacitor to fill up all the space between the electrodes. The mean radius of the electrodes equals `R`, the gap between them is equal to `d`, with `d lt lt R`. The constant voltage `V` is applied across the electrodes of the capacitor. Find the magnitude of the electric force pulling the dielectric into the capacitor. |
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Answer» We known that energy of a capacitor , `U = (q^(2))/(2C)` Hence, from `F_(x) = (del U)/(del x):|_(q =const)`, We have, `F_(x) = (q^(2))/(2) (del c)/(del x)/C^(2)` ....(1) Now, since `d lt lt R`, the capacitance of the given capacitor can be calculated by the formula of a parallel plate capacitor. Therefore, if the dielectric is intoduced upto a depth `x` and the length of the capacitor is `l`, we have, `C = (2pi epsilon_(0) epsilon R x)/(d) + (2pi R epsilon_(0) (l - x))/(d)` ....(2) From (1) and (2) we get, `F_(x) = epsilon_(0) (epsilon - 1) (pi R V^(2))/(d)` |
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| 53. |
The radii of spherical capacitor electrodes are equal to `a` and `b`, with `a lt b`. The interlectordes `epsilon` and resistivity `rho`. Inititally the capacitor is not charged. At the moment `t = 0` the internal electorde gets a charge `q_(0)` Find: (a) the times variation of the charge on the internal elecrtordes, (b) the amount of the heat generated during the spreading of the charge. |
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Answer» (a) Let us mentally isolatate a thin spherical layer with inner and outer radii `r` and `r + dr` respective. Lince of current at all the points of this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness `dr` and cross sectional area `4pi r^(3)`. Now we know that resistance, `dR = rho (dr)/(S(r)) = rho (dr)/(4pi r^(2))` ....(1) Intergating expression (1) between teh Hints, `int_(0)^(R) dR = int_(R)^(b) rho (dr)/(4pi r^(2))` or, `R = (rho)/(4pi) [(1)/(a) - (1)/(b)]` ......(2) Capacitance of the network, `C = (4pi epsilon_(0) epsilon)/([(1)/(a) - (1)/(b)])` ......(3) and `q = C varphi` [where `q` is the charge at any arbitary moment] ....(4) also, `varphi = ((-dq)/(dt)) R`as capacitor is discharging, ....(5) From Eqs.(2), (3),(4) and (5) we get, `q = (4pi epsilon_(0) epsilon)/([(1)/(a) - (1)/(b)]) ([- (dq)/(dt)] rho [(1)/(a) - (1)/(b)])/(4pi)` or, `(dq)/(q) = (dt)/(rho epsilon epsilon_(0))` Intergating `int_(q_(0))^(q) - (dq)/(q) = (1)/(rho epsilon_(0) epsilon) int_(0)^(t) dt = (dt)/(rho epsilon epsilon_(0))` Hence `q = q_(0) e^((-t)/(rho epsilon_(0) epsilon))` (b) From energy conservation heat generated,m during the spreading of the charge, `H = U_(i) - U_(f)` [because `A_("cell") = 0`] `= (1)/(2) (q_(0)^(2))/(4pi epsilon_(0) epsilon) [(1)/(a) - (1)/(b)] - 0 = (q_(0)^(2))/(8pi epsilon_(0) epsion) (b-a)/(ab)` |
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| 54. |
A cylindrical capacitor conneced to a `dc` voltage source `V` touches the surface of water with its end (Fig) The sepration `d` between the capacitors electrodes is substantially less than their mean radius. Find a height `h` to which the water level in the gap will rise. The, capilary effects are to be negelected. |
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Answer» If `C_(0)` is the inital capacitance of the condenser before water rises in it then `U_(i) = (1)/(2) C_(0) V^(2)`, where `C_(0) = (epsilon_(0) 2l pi R)/(d)` (`R` is the mean radius and `l` is the length of the capacitor plates.) Suppose the liquid rises to a height `h` in it. Then the capacitance of the condenser is `C = (epsilon epsilon_(0)h 2pi R)/(d) + (epsilon(l-h) 2pi R)/(d) = (epsilon_(0) 2pi R)/(d) (l + (epsilon - 1) h)` and energy of the capacitor and the liquid (inculding both gravitational and electrostatic condtributions) is epsilon `(1)/(2) (epsilon_(0) 2pi R)/(d) (l + (epsilon - 1)h) V^(2) + rho g (2pi R hd) (h)/(2)` If the capacitor were not connected to a battery this energy would have to be minimized. But the capacitor is connected to the battery and in effect the potential energy of the capacitor and the liquid increases by `delta h ((epsilon_(0) 2pi R)/(2d) (epsilon - 1) V^(2) + rho g (2pi Rd) h)` and that of the cell diminishes by the quantity `A_(cell)` which is the product of charge flown and `V` `delta h (epsilon_(0) (2pi R))/(d) (epsilon - 1) V^(2)` In equilbrium, the two must balance, so `rho g dh = (epsilon_(0) (epsilon - 1) V^(2))/(2d)` Hence `h = (epsilon_(0) (epsilon - 1) V^(2))/(2rho g d^(2))` |
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| 55. |
Inside a parallel-plate capacitor there is a plate parallel to the outer plates, whose thickness is equal to `eta = 0.60` of the gap width. When the plates is absent the capacitor cpaacitance equals `c = 20 muF` . First, the capacitor was connected in parallel to a constant voltage source producting `V = 200 V`, then it was disconnected for it, after which the plates was slowly removed from teh gap. FInd the work perfomed during the removel, if the plate is (a) made of metal, (b) made of glass. |
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Answer» (a) When metal plate of thickness `eta d` is inserted inside the capacitor, capacitance of the system becomes `C_(0) = (epsilon_(0) S)/(d (1 - eta))` Now, intially charge on the capacitor, `q_(0) = C_(0) V = (epsilon_(0) S V)/(d (1 - eta))` Finally, capacitance of the capacitor, `C = (epsilon_(0) S)/(d)` As the source is disconnected, charge on the plates will remain same during the process. Now, from energy conservation, `U_(f) - U_(i) = A_("agent")` (as cell does no work) or, `(1)/(2) (q_(0)^(2))/(C) - (1)/(2) (q_(0)^(2))/(C_(0)) = A_("agent")` Hence `A_("agent") = (1)/(2) [(epsilon_(0) S V)/(d (1 - eta))]^(2) [(1)/(C) - ((1 - eta))/(C)] = (1)/(2) (C V^(2) eta)/((1 - eta)^(2)) = 1.5 mJ` (b) Initally, capacitance of the system is given by : `C_(0) = (C epsilon)/(eta (1 - epsilon) + epsilon)` (this is the capacitance of two capacitors in series) So, charge on the plate , `q_(0) = C_(0) V` Capacitance of the capacitor, after the glass plate has been removed equals `C` from energy conservation, `A_("agent") = U_(f) - U_(i)` `= (1)/(2) q_(0)^(2) [(1)/(C) - (1)/(C_(0))] = (1)/(2) (C V^(2) epsilon eta (epsilon - 1))/([epsilon - eta (epsilon -1)]^(2)) = 0.8 mJ` |
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| 56. |
Each plate of a parallel -plate air capacitor has an area `S`. What amount of work has to be performed to slowly increases teh distance between the plates from `x_(1)` to `x_(2)` if (a) Capacitance of the capacitor which is equal to `q`, or (b) the voltage across the capacitor, which is equal to `V`, is kept constant in the process? |
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Answer» (a) Sought work is equivalent to the work is equavlaent to the work performed against the electric field created by one plate, holding at rest and to bring the other plate away. Therefore the requried work, `A_("agent") = q E(x_(2) - x_(1))`. where `E = (sigma)/(2 epsilon_(0))` is the intensity of the field created by one plate at the location of other. So, `A_("gent") = q (sigma)/(2 epsilon_(0)) (x_(2) - x_(1)) = (q^(2))/(2 epsilon_(0) S) (x_(2) - x_(1))` Alternate : `A_("ext") = Delta U` (as field potential) `= (q^(2))/(2 epsilon_(0)S) x_(2) = (q^(2))/(2 epsilon_(0) S) x = (q^(2))/(2 epsilon_(0) S) (x_(2) - x_(1))` (b) When voltage is kept const, the force acting on each plate acting on each plate of capacitor will depend on the distance between the plates. So, elementary work done by agent, is it displacement over a distanace `dx`, relative to the other, `dA = -F_(x) dx` But, `F_(x) = - ((sigma (x))/(2 epsilon_(0))) S sigma(x)` and `sigma(x) = (epsilon_(0)V)/(x)` Hence, `A = int DA = int_(x_(1))^(x_(2)) (1)/(2) epsilon_(0) (S V^(2))/(x^(2)) dx = (epsilon_(0) S V^(2))/(2) [(1)/(x_(1)) - (1)/(x_(2))]` Alternate : From energy Conservation, `U_(f) - U_(i) = A_(cell) + A_("agent")` or `(1)/(2) (epsilon_(0) S)/(x_(2)) V^(2) - (1)/(2) (epsilon_(0) S)/(x_(1)) V^(2) = [(epsilon_(0) S)/(x_(2)) - (epsilon_(0) S)/(x_(1))] V^(2) + A_("agent")` `(as A_("cell") = (q_(f) - q_(i)) V = (C_(f) - C_(i)) V^(2))` So `A_("agent") = (epsilon_(0) S V^(2))/(2) [(1)/(x_(1)) - (1)/(x_(2))]` |
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| 57. |
There is an infinite straight chain of alternating charges `q` and `-q` . The distance between teh neighbouring charges is equal to `a`. Find the interaction energy of each charge with all the others. Instruction . Make use of the expansion of In `(1 + alpha)` in a power series in `alpha`. |
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Answer» As the chain is of infinite length any two charge of same sign will occur symmetrically to any other charge of opposite sign. So, interaction energy of each chagre with all the others, `U = -2 (q^(2))/(4pi epsilon_(0) a) [1 - (1)/(2) + (1)/(3) - (1)/(4) + .....` up to `oo`] ....(1) But In `(1 + x) = x - (1)/(2) x^(2) + (1)/(3) x^(3)`..... up to `oo` and putting `x = 1` we get IN `2 = 1 - (1)/(2) + (1)/(3) + ....` up to `oo` .......(2) From Eqs (1) and (2), `U = (-q^(2) In 2)/(4pi epsilon_(0) a)` |
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| 58. |
since the period fo revolution of electrons in a unifrom magnetic field rapidly increases with the growth of energy, a cyclotron in unsuitable for their accelearation. This drawback is rectified in a microton (Fig) in which a change `Delta T` in the period of revolution of an electron is made multipile with the period of acclerating field `T_(0)`. How many times has an electron to cross the accerating gap of a microtron to acquire an energy `W = 4.6 MeV` if `Delta T = T_(0)` the magnetic induction is equal to `B = 107 mT`, and the frequency of accelerating field to `v = 3000 MHz` ? |
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Answer» In the nth oribit, `(2pi r_(n))/(v_(n)) = n T_(0) = (n)/(v)`. We igonre the rest mass of the electron and write `v_(n) = c`. Also `W = cp = cBer_(n)`. Thus, `(2pi W)/(B ec^(2)) = (n)/(v)` or, `n = (2pi W v)/(Be c^(2)) = 9` |
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| 59. |
Up to what values of kinetic energy does the period of revolution of an electron and a proton in a uniform magnetic field exceed that at non-relativistic velocities by `eta = 1.0 %` |
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Answer» From (3.279), `T = (2pi epsilon)/(c^(2) eB)` (relativistic). `T_(0) = (2pi m_(0) c^(2))/(c^(2) eVB)` (nonrelativistic), Here, `m_(0) c^(2)//sqrt(1 - v^(2)//c^(2)) = E` Thus, `delta T = (2pi T)/(c^(2) eB), (T = K.E.)` Now, `(delta T)/(T_(0)) = eta = (T)/(m_(0) c^(2))`, so, `T = eta m_(0) c^(2)` |
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| 60. |
(a) An electron moves along a circle of radius 1m in a perpendicular magnetic field of strength 0.50 T. What would be its speed? Is it reasonable?(b) If a proton moves along a circel of the same radius in the same magnetic field, what would be its speed? |
| Answer» Correct Answer - A::B::D | |
| 61. |
A proton is projected with a velocity of `3xx10^(6) m//s` perpendicular to a uniform magnetic field of `0.6 T`.Find the acceleration of the proton.mass of proton`=5/3xx10^(-27) kg` |
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Answer» Correct Answer - A::B::D `f=qVB rArra=f/m=(qVB)/m=864/5xx10^(12) m//s^(2)` |
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| 62. |
A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of radius 4.0 cm as shown in . A wire, carrying a current of 2.0 A, is placed perpendiuclar to and intersecting the axis of the cylindreical region. Find the magnitude of the force acting on the wire. |
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Answer» Correct Answer - A::B `F=BiL L=2r` `=Bi(2r)` `=1xx2xx4xx10^(-2)` `=8xx10^(-2) N` |
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| 63. |
An electron having a kinetic energy of `400 eV` circulates in a path of radius `20 cm` in a magnetic field.Find the magnetic field and the number of revolutions per second made by the electron. |
| Answer» Correct Answer - A::B::D | |
| 64. |
The material for making permanent magnets should have:A. high retentivity, high coercivityB. high retentivity, low coercivityC. low retentivity, high coercivityD. low retentivity, low coercivity |
| Answer» Correct Answer - A | |
| 65. |
A particle is moving with velocity `bar(v)=hat(i)+3hatj` and it produces an electrostatic field at a point given by `bar(E)=2hat(k)`. It will produce magnetic field at that point equal to (all quantities are in S.I. units and speed of light is `c`)A. `(6hati-2hatj)/c^(2)`B. `(6hati+2hatj)/c^(2)`C. zeroD. can not be determined from the given data |
| Answer» Correct Answer - A | |
| 66. |
Find the potential `V` of an electrostatic field `vec E = a(y hat i + x hat j)`, where `a` is a constant. |
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Answer» `d varphi = vec(E). vec(dr) = a (y dx + x dy) = a d (xy)` On intergating, `varphi = a xy + C` |
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| 67. |
Determine the potentail `varphi(x,y,z)` of an electrostatic field `E = ayl + (ax + bz) J + byk`, where `a` and `b` are constants, `I, j, k `are the unit vectors of the axes `x,y,z`. |
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Answer» Given, again `-d varphi = vec(E) . vec(dr) = (ay vec(i) + (ax + bz) vec(j) + by vec(k)). (dx vec(i) + dy vec(j) + dx vec(k))` `= a(y dx + ax dy) + b(z by + y dz) = ad (xy) + bd(yz)` On intergating `varphi = -(a xy + b yz) + C` |
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| 68. |
What would be the interaction force between two coppersphers, each of mass `1 g`, separated by the distance `1m`, if the total electrons charge in them differted from the total charge of the nuclei by one of part cent? |
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Answer» Total number of atons in the sphere fo mass `1 gm = (1)/(63.54) xx10^(23)` So the total auclear charge `lambda = (6.023xx10^(23))/(63.54) xx1.6xx10^(-19)xx29` Now the charge on the sphere = Total nuclear charge - Total electronic change `= (6.023xx10^(23))/(63.54)xx1.6xx10^(-19)xx (29xx1)/(100) = 4.298xx10^(2) C` Hence force of interaction between these two spheres, `F =(1)/(4pi epsilon_(0)) . ([4.398xx10^(2)]^(2))/(1^(2)) N = 9xx10^(9)xx10^(4) xx19.348 N = 1.74xx10^(15) N` |
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| 69. |
Calculate the ratio of the electrostatic of gravitational inteaction forces between two electrons, between two protons. At what values of the specific cahreg `q//m` of a praticle would these forces become equal (in their absoule values) in the case of interaction of indentical) particles ? |
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Answer» `F_(el)` (for electrons) = `(q^(2))/(4pi epsilon_(0) r^(2))` and `F_(gr) = (gamma m^(2))/(r^(2))` Thus, `(F_(el))/(F_(gr))` for electrons) `= (q^(2))/(4pi epsilon_(0) gamma m^(2))` `= ((1.602xx10^(-19) C)^(2))/(((1)/(9xx10^(9))) xx6.67xx10^(-11) m^(3)//(kg . s^(2)) xx (9.11xx10^(-31) kg)^(2)) = 4xx10^(42)` Similarly `(F_(el))/(F_(gr))` (for proton) `= (q^(2))/(4pi epsilon_(0) gamma m^(2))` `= ((1.602xx10^(-19) C)^(2))/(((1)/(9xx10^(9))) xx6.67xx10^(-11) m^(3)//(kg . s^(2)) xx (1.672xx10^(-27) kg)^(2)) = 1xx10^(36)` For `F_(el) = F_(gr)` `(q^(2))/(4pi epsilon_(0) r^(2))` or `(gamma m^(2))/(r^(2))` or `(q)/(m) = sqrt(4pi epsilon_(0) gamma)` `= sqrt((6.67xx10^(-11) m^(3) (kg-s^(2)))/(9xx10^(9))) = 0.86xx10^(-10) C//kg` |
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| 70. |
Two protons move parallel to each other with an equal velcity `v = 300 km//s`. Find the ration of forces of magentic and electricla of the protons. |
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Answer» Force of magentic interaction `vec(F)_(mag) = e(vec(v) xx vec(B))` Where, `vec(B) = (mu_(0))/(4pi) (e (vec(v) xx vec(r)))/(r^(3))` So, `vec(F)_(mag) = (mu_(0))/(4pi) (e^(2))/(r^(3)) [vec(v) xx (vec(v) xx vec(r))]` `= (mu_(0))/(4pi) (e^(2))/(r^(3)) [(vec(v). vec(r)) xx vec(v) - (vec(v) . vec(v)) xx vec(r)] = (mu_(0))/(4pi) (e^(2))/(r^(3)) (-v^(2) vec(r))` And `vec(F)_("ele") = e vec(E) = e (1)/(4pi epsilon_(0)) (e vec(r))/(|vec(r)|^(3))` Hence, `(|vec(F)_(mag)|)/(|vec(F)_("electric")|) = v^(2) mu_(0) epsilon_(0) = ((v)/(c))^(2) = 1.00xx10^(-6)` |
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| 71. |
Determine the electric field strength vector if the potential of this field depends on x, coordinates as(a) `V = a (x^2 — y^2)` (b) `V = axy` where, a is a constant. |
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Answer» (a) Given, `varphi = a(x^(2) - y^(2))` So, `vec(E) = -vec(grad) varphi = -2a (x vec(i) - y vec(j))` The sought shape of field lines is as shown in fig of answersheet assuming `a gt 0` : (b) Since `varphi = axy` So, `vec(E) = -vec(grad) varphi = -ay vec(j) - ax vec(j)` Plot as shown in fig(b) of answersheet. |
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| 72. |
A point dipole with an electric moment `p` oriented in the positive direction of the z axis si located at the origin of coordinates. Find the projections `E_(z)` and `E_(_|_)` of the electric field strength vector (on the plane perpendicular to the z axis at the poiny `S`. At which points is `E` perpendicular to `p` ? |
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Answer» From the results, obtained in the previous problem, `E_(r) = (2P cos theta)/(4pi epsilon_(0) r^(3))` and `E_(theta) = (p sin theta)/(4pi epsilon_(0) r^(3))` From the given figure, it si clear that, `E_(z) = E_(r) cos theta - E_(theta) sin theta = (p)/(4pi epsilon_(0) r^(3)) (3 cos^(2) theta - 1)` and `E_(_|_) = sin theta+ E_(0) cos theta = (3 p sin theta cos theta)/(4pi epsilon_(0) r^(3))` When `vec(E) _|_ vec(p), |vec(E)| = E_(_|_)` and `E_(x) = 0` So `3 cos^(2) theta = 1` and `cos theta = (1)/(sqrt(3))` Thsu `vec(E_(_|_)) vec(p)` at the points located on teh lateral surface of the lateral surface of the come, having its axis, coinciding with the direction ofz-axis and semi vertex angle `theta = cos^(-1) 1//sqrt(3)` |
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| 73. |
A parallel plate capacitor with area of each plate equal to `S` and the separation between them to `d` is put into a stream of conducting liquid with respectivity `rho`. The liquid moves parallel to the plates with a constant velocity `v`. The whoel system is located in a unifrom magentic field of induction `B`, vector `B` being parallel to the plates are interconnected by means of an exteranal resistance `R`. What amount of power is genrated in that resistance? At what value of `R` is the generated power the highest? What is this highest power equla to ? |
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Answer» Resistance of the liquid between the plates `= (rho d)/(S)` Voltage beween the plates `= Ed = v Bd`, Current through the plates `= (vBd)/(R + (rho d)/(S))` Power generted, in the external resistance `R`, `P = (v^(2) B^(2) d^(2) R)/((R + (rho d)/(S))^(2)) = (v^(2) B^(2) d^(2))/((sqrt(R) + (rho d)/(S sqrt(R)))^(2)) = (v^(2) B^(2) d^(2))/([{R^(1//4) - ((rho)/(S sqrt(R)))^(1//2) }^(2) + 2 sqrt((rho d)/(S))]^(2))` This is maximum when `R = (rho D)/(S)` and `P_(max) = (v^(2) B^(2) Sd)/(4 rho )` |
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| 74. |
There is an infinite wire grid with square cells (Fig). The resistance of each wire between neighbouring joint connections is equal to `R_(0)` . Find the resistance `R` of the whole grid between points `A` and `B`. Instruction. Make use of principles of symmetry and superposition. |
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Answer» Suppose that the voltage `V` is applied between the points `A` and `B` then `V = IR = I_(0) R_(0)` where `R` is resistance of whole the grid `I`, the current through the grid and `I_(0)` the current through the segment `AB`. Now from symmetry , `I//4` is the part of the current, flowing through all the four wire segments, meeting at the poin `A` and similarly the amount of current flowing through the wires, meeting at `B` is also `I//4`. Thus a current `1//2` flows through the conductor `AB`, ie. `I_(0) = (1)/(2)` Hence, `R = (R_(0))/(2)` |
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| 75. |
Calculate the time constant `tau` of a straight solenoid of length `l = 1.0 m` having a single-layer winding of copper wire whose total mass is equal to `m = 1.0 kg`. The cross-sectional diameter of the solenoid is assumed to be considerably less than its length. |
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Answer» The time constant `tau` given by `tau = (L)/(R) = (L)/(rho_(0) (l_(0))/(S))` where, `rho_(0)` = resistivity, `l_(0)` = length of the winding wire, `S` = cross section of the wire. But `m = l rho_(0) S` So eliminating `S,tau = (L)/((rho_(0) l_(0))/(m//rho l_(0))) = (mL)/(rho rho_(0) l_(0)^(2))` From problem `3.315 l_(0) = sqrt((4pi r l L)/(mu_(0)))` (note the interchange of `l` and `l_(0)` because of difference in notation here.) Thus, `tau = (mL)/(rho rho_(0) (4pi)/(mu_(0)) L l) = mu_(0) 4pi (m)/(rho rho_(0) l) = 0.7 ms` |
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| 76. |
In the formula `X=3YZ^(2), X` and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are …………………… |
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Answer» Correct Answer - B::C::D `x=3yz^(2)` `[y]=[[x]]/[z]^(2)=(Q^(2)M^(-1)L^(-2)T^(2))/[MQ^(-1)T^(-1)]^(2)=M^(-3)L^(-2)T^(4)Q^(4)` |
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| 77. |
A current `I ` flows radius along a lengthy thin-walled tube of radisu `R` with longiitual slit of width `h`. Find the induction of the magnietic field inside the tube under the condition `h lt lt R` |
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Answer» The thin walled tube with a longitudinal slit can be considered equivalent to a fill tube and a strip carrying the same current density in the opposite direction. Inside the tube, the former does not contibute so the total magnetic field is simply that due to the strip. It is `B = (mu_(0))/(2pi) ((I//2 pi R) h)/(r) = (mu_(0) I h)/(4 pi^(2) R r)` where `r` is the distance of the find point from the strip. |
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| 78. |
A small current-carrying loop is located at a distance `r` from a long straight conductor with current `I`. The magnetic moment of the loop is equal to `p_(m)`. Find the magnitude and direction of the force vector applied to the loop if the vector `p_(m)` (a) is parallel to the stratight conductor, (b) is oriented along the radius vector `r`, (c) coincides in direction with the magnetic field produced by the current `I` the point where the loop is located. |
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Answer» Due to the straight conductor, `B_(varphi) = (mu_(0) I)/(2pi r)` We use the formula, `vec(F) = (vec(p_(m)). vec(grad)) vec(B)` (a) The vector `vec(p_(m))` is parallel to the staright conductor. `vec(F) = p_(m) (del)/(del Z) vec(B) = 0`, because netiher the direction nor the magnitude of `vec(B)` depends on `z` (b) The vector `vec(p_(m))` is oriented along the radius vector `vec(r)` `vec(F) = p_(m) (del)/(del r) vec(B)` The direction of `vec(B)` at `r + dr` is parallel to the direction at `r`. Thus only the `varphi` component of `vec(F)` will survive. `F_(varphi) = p_(m) (del)/(del r) (mu_(0) I)/(2pi r) = - (mu_(0) I p_(m))/(2pi^(2))` (c) The vector `vec(p_(m))` coincides in direction with the magnetic field, produced by the conductor carrying current `I` `vec(F) = p_(m) (del)/(r del varphi) (mu_(0) I)/(2pi) vec(e_(varphi)) = (mu_(0) I p_(m))/(2pi r^(2)) (del vec(e_(varphi)))/(del varphi)` So, `vec(F) = (mu_(0) I p_(m))/(2pi r^(2)) vec(e_(r))` As, `(del vec(e_(varphi)))/(del_(varphi)) = -vec(e_(r))` |
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| 79. |
Define magnetic susceptibility of a material.Name two elements one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? |
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Answer» Magnetic susceptibility:Magnetic susceptibility of material is defined as the ratio of the intensity of magnetisation `(I)`induced in the material to the magnetisation force `(H)` applied on it. Magnetic susceptibility is represented by `X_(m)=I/H` Diamagnetic substances like copper, lead etc. has negative susceptiblity. Paramagnetic substance like aluminimum calcuim etc.has positive susceptibility. Negative susceptibility of diamagnetic substance does not change with temperature. |
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| 80. |
Figure shows a part of an electric circuit. The wires AB, CD, and EF are long and have identical resistances. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistances and the ammeter reads 30 A . Calculate the magnetic force per unit length of AB and CD. ` |
| Answer» Correct Answer - A::C::D | |
| 81. |
Two long straight parallel conductors are separated by a distance of 5 cm and carrying current 20 A. what work per unit length of a conductor must be done to increases the separation between conductors to 10 cm, if the current flows in the same direction ? |
| Answer» Correct Answer - A::B::D | |
| 82. |
What is the direction of the force acting on a charged particle q, moving with a velocity `vec(v)` a uniform magnetic field `vec(B)` ? |
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Answer» `becausevecF=q(vecvxxvecB)` Magnetic force is always perpendicular to magnetic field. |
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| 83. |
A straight , long wire carries a current a current of 20A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10cm of the second wire is `2.0 xx 10^-5 N`, what is the separation between them? |
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Answer» Correct Answer - C::D `(mu_(0)i_(1)i_(2))/(2pid)=F/I` `rArr=2xx10^(-7)xx(20xx20)/d=2xx10^(-4)=2xx10^(-4) rArr d=40cm`. |
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| 84. |
An electrons is moving along `+ve` `x`-axis in the presence of uniform magnetic field along `+ve` `y`-axis. What is the direction of the force acting on it? |
| Answer» The direction of the force is along `-ve Z`-axis. | |
| 85. |
A hydrogen ion of mass m and charge q travels with a speed v along a circle of radius r in a uniform magnetic field of flux density B. Obtain the expression for the magnetic force on the ion and determine its time period. |
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Answer» Magnetic force on the hydrogen ion=Centripetal force or `qvB sin 90^(@)=(mv^(2))/r` or `r=(mv)/(qB)` Time period `T=(2pir)/v=(2pi)/v.(mv)/(qB)=(2pim)/(qB)` |
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| 86. |
Two long parallel straight wires X and Y separated by a distance `5cm` in air carry currents of `10A` and `5A` respectively in opposite directions. Calculate the magnitude and direction of force on a `20cm` length of the wire Y. Figure |
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Answer» `F=(mu_(0)I_(1)I_(2)l)/(2pir)=(4pixx10^(-7)xx10xx5xx0.20)/(2pixx0.05)` `=4xx10^(-5)N` The direction of the force is perpendicular to the length of wire `Y` and acts away from `X`(repulsion) |
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| 87. |
A thin straight horizontal wire of length `0.2 m` whose mass is `10^(-4) kg` floats in a magnetic induction field when a current of `10 ampere` is passed throught it.To make this possible, what should be minimum magnetic strength? |
| Answer» Correct Answer - A::D | |
| 88. |
A `U`-shaped wire of mass `m` and length `l` is immersed with its two ends in mercury (see figure).The wire is in a homogeneous field of magnetic induction `B`.If a charge, that is a current pulse `q=int idt`, is sent through the wire, the wire will jump up.The amount of charge the wire reaches a hight is `Nsqrt15C` assuming that the time of the current pulse is very small in comparionson with the time of flight.Make use of the fact that impulse of force equals `int F dt` which equals `mv`.`B=01(Wb)/m^(2),m=10gm,l=20cm&h=3meters`.Then find the value of `N`. |
| Answer» Correct Answer - A | |
| 89. |
The magnetic field existing in a region is given by `B=B_(0)(1-x/y)hatk`, where `B_(0)` and `l` ae constant `x` is the `x` coordinate of a point and `hatk` is the unit vector along `Z` axis.A square loop of edge and carrying a current is placed with his edges parallel to the `x`,`y` axes.Find the magnitude of the net magnetic force experienced by the loop. |
| Answer» Correct Answer - B | |
| 90. |
A uniform rod of length `L` and mass `M` is hinged at its upper point and is at rest at that moment in the vertical plane.A current `i` flows in it.A uniform magnetic field of strength `B` exists perpendicular to the rod and in horizontal direction is `(iLB)/N`.Then find the value of `N`. |
| Answer» Correct Answer - D | |
| 91. |
Using the expression for volume density of magnetic eneregy, demonstare that the amount of work contributed to magnetization of a unit volume of para or diamagentic, is equal to `A = -JB//2`. |
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Answer» The total energy of the magnetic field is, `(1)/(2) int (vec(B).vec(H)) dV = (1)/(2) int vec(B). ((vec(B))/(mu_(0)) - vec(j)) dV` `= (1)/(2 mu_(0)) int vec(B).vec(B)dV - (1)/(2) int vec(J).vec(B) dV` The secound tern can be interpreted as the energy fo magnetization, and has the density `-(1)/(2) vec(J).vec(B)`. |
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| 92. |
How much (in per cent) has a filament diameter decreased due to evaporation if the maintances of the previous temperature due to evaporation if the maintenance of the previous temperatur required an increases of the voltage by `eta = 1.0 %` ? The amount of heat transfereed fromt the filament into surrounding space is assumed to be propotional to the filament surface area. |
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Answer» If the wire diameter decreases by `sigma` then by the information given `P` = Power input `= (V^(2))/(R)` = beat lostg throgh the surface, `H`. Now, `H prop (1 - sigma)` like the surface area and `R prop (1 - sigma)^(2)` So, `(V^(2))/(R_(0)) (1 - sigma)^(2) = A (1 - sigma)` or, `V^(2) (1 - sigma)` = constant But `V prop 1 + eta` so `(1 + eta)^(2) (1 - sigma)` = constant = 1 Thus `sigma = 2 eta = 2%` |
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| 93. |
A non-relativistic charged particle files through the electric field of a cyclindrical capacitor and gets into a unifrom transverse magnetic field with induction `B`(fig). In the capacitor the particle moves along the are of a circle, in the magnetic field, along a semi-circle of radius `r`. The potential differnce applied to the capacitor is equal to `V`, the radii of the electrodes are equal to `a` and `b`, with `a lt b`. Find the velocity of the particle and its specific charge `q//m`. |
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Answer» Inside the capacitor, the electric field follows `a (1)/(r)` law, and so the potential can be written as `varphi = (V In r//a)/(In b//a), E = (-V)/(In b//a) (1)/(r)`, Here `r` is the distance from the axis of the capacitor. Also, `(mv^(2))/(r) = (qV)/(In b//a r) (1)/(r)` or `mv^(2) = (qV)/(In b//a)` On the other hand `mv = q B r` in the magentic field. Thus, `v = (V )/(B r In b//a)` and `(q)/(m) = (v)/(Br) = (V)/(B^(2) r^(2) In (b//a))` |
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| 94. |
The ill effects associated with the variatiion of the period of revolution of the particle in a cyclotron due to the increase of its energy are eleminated by slow monitoring (modulating ) the frequency of accelerting field. According to what law `omega (t)` should this frequencecy by mointored if the masgnetic induction is equal to `B` and the particle acquires an energy `Delta W` per revolution ? The charge of the particle is `q` and its mass is `m`. |
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Answer» The basic condition is the relatistiv equaction, `(mv^(2))/(r) = Bqv`, or, `mv = (m_(0) v)/(sqrt(1 - v^(2)//c^(2))) = Bqr`. Or calling, `omega = (Bq)/(m)`, we get, `omega = (omega_(0))/(sqrt(1 + (omega_(0)^(2) r^(2))/(c^(2)))), omega_(0) = (Bq)/(m_(0)) r` is the radius of the instantanceous orbit, The time of acceleration is, `t = sum_(n = 1)^(N) (1)/(2 v_(n)) = sum_(n = 1)^(N) (pi)/(omega_(n)) = sum_(n)^(N) (pi W_(n))/(q Bc^(2))` `N` is the numbr of crossing of either `Dee`. But,n `W_(n) = m_(0) c^(2) + (n Delta W)/(2)`, there being two crossings of the Dees per revolution. So, `t = sum (pi m_(0) c^(2))/(qB c^(2)) + sum (pi Delta W_(n))/(2q B c^(2))` `= N (Pi)/(omega_(0)) + (N (N + 1))/(4) (pi Delta W)/(qB c^(2)) = N^(2) (pi Delta W)/(4 q Bc^(2)) (N gt gt 1)` Also, `r = r_(N) (v_(N))/(omega_(N)) = (c)/(pi) (del t)/(del N) = (Delta W)/(2q Bc) N` Hence finally, `omega = (omega_(0))/(sqrt(1 + (q^(2) B^(2))/(m_(0)^(2) c^(2)) xx (Del W^(2))/(4 q^(2) B^(2) c^(2))) N^(2))` `= (omega_(0))/(sqrt(1 + ((Delta W)^(2))/(4 m_(0)^(2) c^(4)) xx (4 q Bc^(2))/(pi Delta W) t)) = (omega_(0))/(sqrt(1 + at))`, `a = (q B Delta W)/(pi m_(0)^(2) c^(2))` |
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| 95. |
A particle with specific charge `q//m` is located inside a round solenoid at a distance `r` from its axis. With the current swichted into the winding, the magnetic induction of the field generated by the solenoid amounts to `B`. Find the velocity of the particle and the curvature radius of its trajectory,n assuming that during the increase of current flowing in the solenoid the particle shifts by a negligible distance. |
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Answer» When the magnetic field is being set up in the solenoid, and electric field will be induced in it, this will accelerate the charged paritcle. If `B` is the rate, at which the matgnetic field is increasing, then. `pi r^(2) dotB = 2pi r E` or `E = (1)/(2) r dotB` Thus, `m (dv)/(dt) = (1)/(2) r dotB q`, or `v = (qBr)/(2m)`, After the field is set up, the parcticle will execute a circular motion of radius `rho`, where `mv = B q rho`, or `rho = (1)/(2) r` |
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| 96. |
A long wire carrying `i` is bent to form a plane angle `theta`.Find the magnetic field `B` at a point on the bisector of this angle situated at a distance `x` from the vertex is written in the form of `K cot theta/4`Tesla.Then, find the value of `K`. |
| Answer» Correct Answer - B | |
| 97. |
Two circular coils of wire each having a radius of `4cm` and `10 turns` have a common axis and are `3cm` apart If a current of `1A` passes through each coil in the opposite direction find the magnetic induction (i) At the center of either coil (ii) At a point on the axis, midway between them . |
| Answer» Correct Answer - A::B::C | |
| 98. |
Half of an infinitely long straight current-carrying solenoid is filled with magnetic substance as shown in Fig. Draw the appoximater plots of magnetic induction `B`, strength `H` and magnetization `J` on the axis as functions of `x`. |
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Answer» We can obtain the form of the curves, required here, by quanlitative arguments. From `oint vec(H).d. vec(l) = I` we get `H(x gt gt o) = H(x lt lt o) = nl` Then `B(x gt gt 0) = mu mu_(0) nI` `B(x lt lt o) = mu_(0) nI` Also, `B (x lt 0) = mu_(0) H (x lt o)` `J(x lt 0) = 0` `B` is continous at `x = 0`, `H` is not. These give the required curves as shwon in the anser sheet. |
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| 99. |
An infinitely long straight wire carries a current `i_(1)`,as shown in the figure.The total force on another wire `CD` of length `l` which is placed so that `c` is at a distance a from the current carrying wire is `(mu_(0)i_(1)i_(2))/(Npi)ln((a^(2)+L^(2))/a^(2))`.Then find value of `N` |
| Answer» Correct Answer - D | |
| 100. |
A condutor carrying current `I` is placed parallel to a current per unit width `j_(0)` and width `d`, as shown in the Find the force per unit length on the conductor . |
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Answer» Correct Answer - B Field due to strip `B=(mu_(0)j_(0))/pi tan^(-1)(d/(2h))` towards `x` axis Force on unit length `dl` of wire `dvecF=i[dlhatjxxBhati]` `rArr (dF)/(dl)=(mu_(0)j_(0))/pi tan^(-1)(d/(2h))(-hatk)` |
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