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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.
| 1. |
A particle having a charge of `2.0X10^(-8) C` and a mass of `2.0X10^(-10)g` is projected with a speed of `2.0X10^3 ms^(-1)` in a region having a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period. |
| Answer» Correct Answer - A::B::C | |
| 2. |
A particle of mass `m` and charge `q` moves with a constant velocity `v` along the positive `x` direction. It enters a region containing a uniform magnetic field `B` directed along the negative `z` direction, extending from `x = a` to `x = b`. The minimum value of `v` required so that the particle can just enter the region `x gt b` isA. `(qbB)/m`B. `(q(b-a)B)/m`C. `(qzB)/m`D. `(q(b+a)B)/(2m)` |
| Answer» Correct Answer - B | |
| 3. |
A magentic flux through a statinary loop with a resistance `R` varies during the tiem interval `tau` as `Phi = at (tau - t)`. Find the amount of heat generated in the loop during that time. The inductance of the loop is to be neglected. |
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Answer» The amount of heat generated in the loop during a small time interval `dl`, `d Q = xi^(2)//R dt`, but `xi = - (d Phi)/(dt) = 2 at-a tau`, So, `d Q = ((2 at - a tau)^(2))/(R) dt` and hence, the amoung of heat, generated in the loop during the time interval 0 to `tau`. `Q = int_(0)^(tau) ((2at - a tau)^(2))/(R) dt = (1)/(3) (a^(2) tau^(3))/(R)` |
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| 4. |
A direct current of density `j` flows along a round unifrom staright wire with cross-section radius `R`. Find the magnectic induction vector of this current at the point whose position relative to the axis of the wire is defined by a radius vector `r`. The magentic permeabilility is assumed to be equla to unity thoughhout all the space. |
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Answer» By circulation theorem inside the conductor `B_(varphi) 2pi r = mu_(0) j_(x) pi r^(2)`, or `B_(varphi) = mu_(0) j_(x) r//2` i.e., `vec(B) = (1)/(2) mu_(0) vec(j) xx vec(r)` Similarly outside the conductor, `B_(varphi) 2pi r = mu_(0) j_(x) pi R^(2)` or, `B_(varphi) = (1)/(2) mu_(0)j_(x) (R^(2))/(r)` So, `vec(B) = (1)/(2) mu_(0) (vec(j) xx vec(r)) (R^(2))/(r^(2))` |
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| 5. |
A supercondutig round ring of radius `a` and inductance `L` was located in a unifrom magnetic fied of induction `B`. The ring plane was parallel to the vector `B`, and the current in the ring was equal to zero. Them the ring was turned through `90^(@)` so that its plane became perpendicular to the feild. FInd: (a) the current induced in the ring after the turn, (b) the work perfromed during the turn. |
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Answer» In a superconductor there is no resistance, Hence, `L (dI)/(dt) + (d Phi)/(dt)` So intergating `I = (Delta Phi)/(L) = (pi a^(2) B)/(L)` because `Delta Phi = Phi_(f) - Phi_(i), Phi_(f) = pi a^(2) B, Phi_(i) = 0` Also, the work done is `A = int xi I dt = int (d Phi)/(dt) = (1)/(2) L I^(2) = (1)/(2) (pi^(2) a^(4) B^(2))/(L)` |
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| 6. |
A system consists of two parallel planes carrying currents producing a unifrom magnetic field of induction `B` between the planes. Outside this space there is no magnetic field. Find the magnetic force acting this space there is no magentic field. Find the magnetic force acting per unit area of each plane. |
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Answer» By the circulation theorem `B = mu_(0) i`, where `i` = current per unit length flowing along the plane perpendicular to the sphere. Currents flow in the opposite sense in the two planes is just `(1)/(2) B`. The force on the plate in, `(1)/(2) B xx i xx` Length `xx` Breadth `= (B^(2))/(2 mu_(0))` per unit area. (Recall the formula `F = BIl` on a straight wire.) |
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| 7. |
Find the magnetic induction at the point `O` if the wire carrying a current `I = 8.0 A` has the shwon in Fig. The radius of the curved part of the wire is `R = 100 mm`, the linear parts of the wire are very long. |
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Answer» (a) `vec(B_(0)) = vec(B_(1)) + vec(B_(2)) + vec(B_(3))` `= (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) pi (-vec(i)) + (mu_(0))/(4pi) (i)/(R) (-vec(k))` `= -(mu_(0))/(4pi) (i)/(R) [2vec(k) + pi l]` So, `|vec(B_(0))| = (mu_(0))/(4pi) (i)/(R) sqrt(pi^(2) + 4) = 0.30 mu T` (b) `vec(B_(0)) = vec(B_(1)) + vec(B_(2)) + vec(B_(3))` `= (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) pi (-vec(i)) + (mu_(0))/(4pi) (i)/(R) (-vec(i))` `= -(mu_(0))/(4pi) (i)/(R) [vec(k) + (pi + 1) vec(i)]` So, `|vec(B_(0))| = (mu_(0))/(4pi) (i)/(R) sqrt(1 + (pi + 1)^(2)) = 0.34 mu T` (c) Here using the law of parallel resistances `i_(1) + i_(2) = 1` and `(i_(1))/(i_(2)) = (1)/(3)`, So, `(i_(1) + i_(2))/(i_(2)) = (4)/(3)` Hence `i_(2) = (3)/(4) i`, and `i_(1) = (1)/(4) i` Thus `vec(B_(0)) = (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) (-vec(j)) + [(mu_(0))/(4pi) ((3pi)/(2)) (i_(1))/(R) (-vec(i)) + (mu_(0))/(4pi) ((pi//2) i_(2))/(R) vec(i)]` `= (-mu_(0))/(4pi) (i)/(R) (vec(j) + vec(k)) + 0` Thus, `|vec(B_(0))| = (mu_(0))/(4pi) (sqrta(2)i)/(R) = 0.11 mu T` |
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| 8. |
Calculate magnetic induction at point O if the wire carrying a current I has the shape shown in Fig. (The radius of the curved part of the wire is equal to R and linear parts of the wire are very long.) |
| Answer» Correct Answer - A::B::C::D | |
| 9. |
A circular loop of area `1cm^2` , carrying a current of 10 A , is placed in a magnetic field of 0.1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field isA. zeroB. `10^(-4) N-m`C. `10^(-2) N-m`D. `1 N-m` |
| Answer» Correct Answer - A | |
| 10. |
`N = 300` turns of thin wire are uniformly wound on a permanent magnet shaped as a cylinder whose length is equla to `l = 15 cm`. When a current `I = 3.0 A` was passed through the wring the field outside the magnet disappeared. Find the coercive force `H_(e)` of the materail from which the magnet was manufaucured. |
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Answer» The corecive force `H_(e)` is just the magnitude field within the cylinder. This is by circulation theorem, `H_(e) = (N I)/(l) = 6 kA//m` (from `oint vec(H).d.vec(r) = I`, total current considering a recantagular contour.) |
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| 11. |
In Hall effect measurements ina sodium conductor the strength of a transervse field was found to be equla to `E = 5.0 mu V//cm` with a current density `j = 200 A//cm^(2)` and magentic induction `B = 1.00 T`. Find the conentrations of the condiction electrons and its ratio to the total number of atoms in the given conductor. |
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Answer» Here, `v_(d) = (E)/(B)` and `j = "ne" v_(4)` so, `n = (jB)/(eE) = (200xx10^(4) (A)/(m^(2)) xx 1 T)/(1.6xx10^(-19)C xx 5xx10^(-4)) V//m` `= 2.5xx10^(28) per m^(3) = 2.5xx10^(22)` per `c.c.` Atomic weight of `Na` being 23 and its density = 1, motar volume is `23 c.c.` This number of atoms per unit volume is `(6xx10^(23))/(23) = 2.6xx10^(22)` per `c.c`. Thus there is alomost one conduction electron per atom. |
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| 12. |
Find the mobility of the conduction electrons in a copper conductor if in Hall effect measuremaents performed in the magnetic fileld of induction `B = 100 mT` the transverse electric field strength of the given conductor turned out be `eta = 3.1.10^(3) `times less than that of the longtudinal electric field. |
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Answer» By definiation, modility = `("dirft velocity")/("Electric field compon ents causing this drift")` or `mu = (v)/(E_(L))` On other hand, `E_(T) = vB = (E_(L))/(eta)`, as given so, `mu = (1)/(eta B) = 3.2xx10^(-3) m^(2)//(V .s)` |
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| 13. |
A square current carrying loop made of thin wire and having a mass `m=10g` can rotate without friction with respect to the vertical axis `OO_(1)` passing throught the centre of the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneous magnetic field with and induction `B =10^(-1)T` directed at right angles to the plane of the drawing `A` current `I=2A` is flowing in the loop Find the period of small oscillations that the loop performs about its position of stable equilibrium |
| Answer» Correct Answer - A::B | |
| 14. |
A uniform dielectric ball is placed in a unifrom electric fileld of strength `E_(0)`. Under these conditions teh dielectric becomes polarized uniformly. Find the electric field strength `E` inside teh ball and the polarization `P` of the dielectric whose permittively equals `epsilon`. Make use of the reslult obtained in Problem 3.96 |
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Answer» By superposition the field `vec(E)` inside the ball is given by `vec(E) = vec(E_(0)) - (vec(P))/(3epsilon_(0))` On the other hand, if the space is not too small, the macrospic equation `vec(P) = (epsilon - 1) epsilon_(0) vec(E)` 0r, `vec(E) = (3 vec(E_(0)))/(epsilon + 2)` Also, `vec(P) = 3epsilon_(0) (epsilon - 1)/(epsilon + 2) vec(E_(0))` |
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| 15. |
A long round cylinderical made of uniform dielectric is placed in a uniform electric field of strength `E_(0)`. The axis of the cylinder is perpendicular to vector `E_(0)`. The axis of the cylinder is perpendiucular to vector `E_(0)`. Under these conditions the dielectrics becomes polarized unifromly. making use of the result obtained in the foregoing probem, find the electric field strength `E` is the cylinder and the polarization `P` of the distance whose permittivity is equal to `epsilon`. |
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Answer» As in 3.98, we wire `vec(E) = vec(E) = vec(E_(0)) - (vec(P))/(2epsilon_(0))` using being the result of the foregoing problem. Also, `vec(P) = (epsilon - 1) epsilon_(0) vec(E)` So, `vec(E) ((epsilon + 1)/(2)) = vec(E_(0))`, or, `vec(E) = (2 vec(E_(0)))/(epsilon + 1)` and `vec(P) = 2epsilon_(0) (epsilon - 1)/(epsilon + 1) vec(E_(0))` |
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| 16. |
A tightly-wound, long solenoid has n turns per unit length, a radius r and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid? |
| Answer» Correct Answer - B | |
| 17. |
A very long straight solenoid carries a current `i`. The cross-sectional area of the solenoid is equal to `S`, the number of turns per unit length is equal to `n`. Find the flux of the vector `B` through the end plane of the solenoid. |
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Answer» The magnetic field is given by `B = (1)/(2) mu_(0) nI (1 - (x)/(sqrt(x^(2) + R^(2))))` At the end, `B = (1)/(2) mu_(0) n I = (1)/(2) B_(0)`, where `B_(0) = mu_(0) nI`, is the field deep inside the solement. Thus, `Phi = (1)/(2) mu_(0) n IS = Phi//2` where `Phi = mu_(0) nl S` is the flux of the vector `B` through the cross setion deep inside the solenid. |
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| 18. |
A magnet is 10cm long and its pole strength is 120 CGS units (1 CGS unit of pole strength `= 0.1 A m)`. Find the magnitude of the magnetic field `B` at a point on its axis at a distance 20cm from it. |
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Answer» The pole strength is `m=120 CGS units =12A-m` Magnetic length is `2l=10vm ` or `l=0.05 m`. Distance from the magnet is `d=20cm=0.2 m`.The field `B` at a point in end-on position is `B=(mu_(0))/(4pi)(2Md)/((d^(2)-l^(2))^(2)` `=(mu_(0))/(4pi)(4Mld)/((d^(2)-l^(2))^(2)` `=(10^(-7)(T-m)/A)(4xx(12A-m)xx(0.05m)xx(0.2m))/[(0.2m^(2))-(0.05m^(2))]^(2)` `=3.4xx10^(-5) T`. |
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| 19. |
A solenoid of length `0.4m` and having `500` turns of wire carries a current of `3amp`.Calculate the torque requred to hold a coil (having radius `0.02 cm` current `2A` and turns `50`)in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.`(pi^(2)=10)` |
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Answer» Correct Answer - `3/5xx10^(-7) Nm` `B=mu_(0)ni` Torque acting on the second coil `tau=M B sin 90^(@)` `=(N_(2)i_(2)A_(2))B` `=N_(2)i_(2)xxAxxmu_(0)nxxi` `tau=3/5xx10^(-7)Nm` |
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| 20. |
Determine the magnetic field at the centre of the current carrying wire arrangement shown in Fig. The arrangement extends to infinity. (The wires joining the successive squares are along the line passing through the centre). A. `(mu_(0)I)/(sqrt2pia)`B. 0C. `(2sqrt2mu_(0)I)/(pia)"ln" 2`D. none of these |
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Answer» Correct Answer - C `vecB_("due to first loop")=4(mu_(0)i)/(4pi1/2)[cos45^(@)+cos45^(@)]=(2sqrt2mu_(0)i)/(pia)` `vecB_("due to first loop")=(-4mu_(0)i)/((4pi)(2a)/2)[cos45^(@)+cos45^(@)]=(-sqrt2mu_(0)i)/(pia)` `vecB=(2sqrt2mu_(0)i)/(pia)[1-1/2+......oo]=(2sqrt2mu_(0)i)/(pia)` in`2` |
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| 21. |
A horizontal overheadpowerline is at height of `4 m` from the ground and carries a current of `100 A` from east to west. The magnetic field directly below it on the ground is `( nu_(0) = 4 pi xx 10^(-7) Tm A^(-1)`A. `5xx10^(-6)T` northwardB. `5xx10^(-6)T` southwardC. `2.5xx10^(-7)T` northwardD. `2.5xx10^(-7)T` southward |
| Answer» Correct Answer - 2 | |
| 22. |
Relative permitivity and permeability of a material `epsilon_(r) and mu_(r)`, respectively . Which of the following values of these quantities are allowed for a diamagnetic material?A. `epsilon_(r)=1.5,mu_(r)=0.5`B. `epsilon_(r)=0.5,mu_(r)=0.5`C. `epsilon_(r)=1.5,mu_(r)=1.5`D. `epsilon_(r)=0.5,mu_(r)=1.5` |
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Answer» Correct Answer - 1 `epsilon_(r)` is always greater than `1`. |
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| 23. |
A loop carrying current `I` lies in the ` x-y ` plane as shown in the figure . The unit vector ` hat k` is coming out of the plane of the paper . The magnetic moment of the current loop is A. `a^(2)Ihatk`B. `(pi/2+1)a^(2)Ihatk`C. `-(pi/2+1)a^(2)Ihatk`D. `(2pi+1)a^(2)Ihatk` |
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Answer» Correct Answer - B Area`=a^(2)+4xxpi((a)/(2))^(2)/2` `=a^(2)+(pia^(2))/2` `A=(1+pi/2)a^(2)hatk` |
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| 24. |
The figure shows a circular loop of radius `a` with two long parallel wires `(numbered 1 and 2)` all in the plane of the paper . The distance of each wire from the centre of the loop is `d`. The loop and the wire are carrying the same current `I`. The current in the loop is in the counterclockwise direction if seen from above. (q) The magnetic fields(B) at `P` due to the currents in the wires are in opposite directions. (r) There is no magnetic field at `P`. (s) The wires repel each other. (5) Consider `dgtgta`, and the loop is rotated about its diameter parallel to the wires by `30^(@)` from the position shown in the figure. If the currents in the wire are in the opposite directions, the torque on the loop at its new position will be ( assume that the net field due to the wires is constant over the loop).A. `(mu_(0)I^(2)a^(2))/d`B. `(mu_(0)I^(2)a^(2))/(2d)`C. `(sqrt3mu_(0)I^(2)a^(2))/d`D. `(sqrt3mu_(0)I^(2)a^(2))/(2d)` |
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Answer» Correct Answer - B Magnetic field at mid point of two wires `=(mu_(0)l)/(pid)o x` Magnetic moment of loop `=lpia^(2)` Torque on loop `=M B sin 150^(@)` `=(mu_(0)I^(2)a^(2))/(2d)` |
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| 25. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :A. `(mu_(0)l(b-a))/(24ab)`B. `(mu_(0)iI)/(4pi)[(b-a)/(ab)]`C. `(mu_(0)I)/(4pi)[2(b-a)+pi/3(a+b)]`D. zero |
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Answer» Correct Answer - 1 Magnetic field due to loop `ABCD` `(mu_(0)I)/(4pi)(pi/6)xx[1/a-1/b] =(u_(0)I)/24[(b-a)/(ab)]` |
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| 26. |
An electric charge `+q` moves with velocity `vecv=3hati+4hatj+hatk`,in an electromagnetic field given by: `vecE=3hati+hatj+2hatk` and `vecB=hati+hatj+3hatk`.The `y`-component of the force experienced by `+q` is:A. `7 q`B. `5 q`C. `3 q`D. `2 q` |
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Answer» Correct Answer - 1 `vecF=q[vecE+vecvxxvecB]=F_(y)=7q hatj` |
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| 27. |
A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin . Due to the process of the current `I_(1)` at the origin:A. The forces on `AD` and `BC` are zero.B. The magnitude of the net force on the loop is given by `(mu_(0)I_(1)I)/(4pi)[2(b-a)+pi/3(1+b)]`C. The magnitude of the net force on the loop is given by `(mu_(0)II_(1))/(24ab)(b-a)`D. the forces on `AB` and `DC` are zero. |
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Answer» Correct Answer - 1 Magnetic field due to `l_(1)` is parallel to `AD` and `BC`.So that force On `AD` and `BC` is zero. |
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| 28. |
The figure shows a long striaght wire of a circular cross-section (radius a) carrying steady current I.The current I is unifromly distributed across this distance `a//2` and 2a from axis isA. `1//4`B. `4`C. `1`D. `1//2` |
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Answer» Correct Answer - 3 `B_(1)=(mu_(0)i)/(2pia^(2))` where `0lerlea B_(1)=(mu_(0)i)/(2pia^(2))a/2 (at r=a/2)` `(mu_(0)i)/(4pia^(2)),B_(2)=(mu_(0)i)/(2pi(2a)) (at r=2a) B_(1)/B_(2)=1` |
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| 29. |
In the given figure of a cyclotron, showing the particle source `S` and the dees.A uniform magnetic field is directed up form the plane of the page.Circulating protons spiral outward within the hollow dees gaining energy every time they cross the gap between the dees.Suppose that a proton, injected by source `S` at the centre of the centre of the cyclotron in figure initially moves toward a negatively charged dee.It will accelerate toward this dee and enter it.Once inside, it is shielded from electric field by the copper walls of the dee,that is the electric field does not enter the dee.The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in circular path whose radius, which depends on its speed, is given by `r=(mv)/(qB)`...(1) Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed.Thus, the proton again faces a negatively charged dee and is again accelerated.Thus,the proton again faces a negatively charged dee and is again accelerated.This process continues, the circulating proton always being in step.with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system.There a deflector plate sends it out through a portal.The key to the operation of the cyclotron is that the frequency `f` at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency `f_(osc)` of the electrical oscilliator, or `f=f_(osc)("resonance condition")`...(2) This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency `f_(osc)` that is equal to the natural frequency `f` at which the proton circulates in the magnetic field. Combining equation `1` and `2` allows us to write the resonance condition as `qB=2pimf_(osc)`..(3) For the proton, `q` and `m` are fixed.The oscillator (we assume) is designed to work at a single fixed frequency `f_(osc)`We then "tune" the cyclotron by barying `B` until eq. `3` is satisfied and then many protons circulate through the magnetic field, to emerge as a beam. Ratio of radius of successive semi circular path A. `sqrt1:sqrt2:sqrt3:sqrt4..`B. `sqrt1:sqrt2:sqrt5..`C. `sqrt2:sqrt4:sqrt6..`D. `1:2:3….` |
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Answer» Correct Answer - B When charge is accelerated by electric field it gains energy for first time `KE_(1)=(qV)/2` for second time `KE_(2)=3/2qV` for third time `KE_(3)=5/2qV` hence the ratio of radii are `r_(1):r_(2):r_(3): ....::sqrt(2mqv)/(qB):sqrt(2m3qv)/(qB): ...` `r_(1):r_(2):r_(3): ....::sqrt1:sqrt3:sqrt5 ...` |
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| 30. |
In the given figure of a cyclotron, showing the particle source `S` and the dees.A uniform magnetic field is directed up form the plane of the page.Circulating protons spiral outward within the hollow dees gaining energy every time they cross the gap between the dees.Suppose that a proton, injected by source `S` at the centre of the centre of the cyclotron in figure initially moves toward a negatively charged dee.It will accelerate toward this dee and enter it.Once inside, it is shielded from electric field by the copper walls of the dee,that is the electric field does not enter the dee.The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in circular path whose radius, which depends on its speed, is given by `r=(mv)/(qB)`...(1) Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed.Thus, the proton again faces a negatively charged dee and is again accelerated.Thus,the proton again faces a negatively charged dee and is again accelerated.This process continues, the circulating proton always being in step.with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system.There a deflector plate sends it out through a portal.The key to the operation of the cyclotron is that the frequency `f` at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency `f_(osc)` of the electrical oscilliator, or `f=f_(osc)("resonance condition")`...(2) This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency `f_(osc)` that is equal to the natural frequency `f` at which the proton circulates in the magnetic field. Combining equation `1` and `2` allows us to write the resonance condition as `qB=2pimf_(osc)`..(3) For the proton, `q` and `m` are fixed.The oscillator (we assume) is designed to work at a single fixed frequency `f_(osc)`We then "tune" the cyclotron by barying `B` until eq. `3` is satisfied and then many protons circulate through the magnetic field, to emerge as a beam. Distance travelled in each time period are in the ratio of: A. `sqrt1+sqrt3:sqrt5+sqrt7:sqrt9+sqrt11`B. `sqrt2+sqrt3:sqrt4+sqrt5:sqrt6+sqrt7`C. `sqrt1:sqrt2:sqrt3`D. `sqrt2:sqrt3:sqrt4` |
| Answer» Correct Answer - A | |
| 31. |
In the given figure of a cyclotron, showing the particle source `S` and the dees.A uniform magnetic field is directed up form the plane of the page.Circulating protons spiral outward within the hollow dees gaining energy every time they cross the gap between the dees.Suppose that a proton, injected by source `S` at the centre of the centre of the cyclotron in figure initially moves toward a negatively charged dee.It will accelerate toward this dee and enter it.Once inside, it is shielded from electric field by the copper walls of the dee,that is the electric field does not enter the dee.The magnetic field, however, is not screened by the (nonmagnetic) copper dee, so the proton moves in circular path whose radius, which depends on its speed, is given by `r=(mv)/(qB)`...(1) Let us assume that at the instant the proton emerges into the center gap from the first dee, the potential difference between the dees is reversed.Thus, the proton again faces a negatively charged dee and is again accelerated.Thus,the proton again faces a negatively charged dee and is again accelerated.This process continues, the circulating proton always being in step.with the oscillations of the dee potential, until the proton has spiraled out to the edge of the dee system.There a deflector plate sends it out through a portal.The key to the operation of the cyclotron is that the frequency `f` at which the proton circulates in the field (and that does not depend on its speed) must be equal to the fixed frequency `f_(osc)` of the electrical oscilliator, or `f=f_(osc)("resonance condition")`...(2) This resonance condition says that, if the energy of the circulating proton is to increase, energy must be fed to it at a frequency `f_(osc)` that is equal to the natural frequency `f` at which the proton circulates in the magnetic field. Combining equation `1` and `2` allows us to write the resonance condition as `qB=2pimf_(osc)`..(3) For the proton, `q` and `m` are fixed.The oscillator (we assume) is designed to work at a single fixed frequency `f_(osc)`We then "tune" the cyclotron by barying `B` until eq. `3` is satisfied and then many protons circulate through the magnetic field, to emerge as a beam. Change in kinetic energy of charge particle after every time period is: A. `2qV`B. `qV`C. `3qV`D. none of these |
| Answer» Correct Answer - A | |
| 32. |
A charged particle goes undeflected in a region containing uniform electric and magnetic field.It is not possible at all `(vecupsilon="velocity of particle", vecE="uniform electric field", vecB="uniform magnetic field")`A. `vecE||vecB.vecupsilon||vecE`B. `vecE` is not collinear to `vecB`C. `vecupsilon||vecB`but `vecE` is not collinear to `vecB`D. `vecE||vecB`but `vecupsilon` is not collinear to `vecE` |
| Answer» Correct Answer - C,D | |
| 33. |
An electron accelerated by a potnetial difference `V = 1.0 kV` moves in a unifrom magentic field at angle `alpha = 30^(@)` to the vector `B` whose modulus is `B = 29 mT`. Find the pithch of the helical trajectroy of the electron. |
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Answer» `T = eV = (1)/(2) mv^(2)` (The given potential difference is not large enough to cause significant deviations from the nonletivistic formula). Thus,. `v = sqrt((2eV)/(m))` So, `v_(||) = sqrt((2eV)/(m)) cos alpha, v_(_|_) = sqrt((2eV)/(m)) sin alpha` Now, `(m v_(_|_)^(2))/(r) = B ev_(_|_)` or, `r = (m v_(_|_))/(Be)`, and `T = (2pi)/(v_(_|_)) = (2pi m)/(Be)` Pitch `p = v_(||) T = (2pi m)/(Be) sqrt((2eV)/(m)) cos alpha = 2pi sqrt((2mV)/(eB^(2))) cos alpha` |
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| 34. |
Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The ring is rotated about its axis with an angular speed `omega` Show that the magnetic moment `mu` and the angular momentum I of the plate are rlated as `mu = (q)/(2m)l.` |
| Answer» Correct Answer - A::B | |
| 35. |
An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field ? |
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Answer» The direction of magnetic field `B` is parallel or antiparallel to the velocity `vecV` of electron. As `F=q(vecvxxvecB)=0 "since"vecv||vecB` |
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| 36. |
Which of the following relations is not correct?A. `B=mu_(0)(H+I)`B. `B=mu_(0)H(1+chi_(m))`C. `mu_(0)H(1+chi_(m))`D. `mu=1+chi_(m)` |
| Answer» Correct Answer - C | |
| 37. |
When a ferromagnetic material goes through a hysteresis loop, the magnetic susceptibilityA. has a fixed valueB. may be zeroC. may be infiniteD. may be negative |
| Answer» Correct Answer - A | |
| 38. |
The hysteresis cylcle for the material of a permanent magnet isA. short and wideB. tall and norrowC. tall and wideD. short and narrow |
| Answer» Correct Answer - C | |
| 39. |
Two metal balls of the same radius `a` are located in a homongenous poorly conducting medium with resistivity `rho`. Find the resistance of the medium between the balls provided that the separation between them is mush greater than the radius of the ball. |
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Answer» Let us mentally impart the charge `+q` and `-q` to the balls respectively. The electric field strength at the surface of a ball will be determined only by is own charge and the charge can be considred to be uniformly distributed over the surface, because the other ball is at infinite distance. Magnitude of the field strength is given by, `E = (q)/(4pi epsilon_(0) a^(2))` So, current density `j = (1)/(rho) (q)/(4pi epsilon_(0) a^(2))` and electric current `I = int vec(j). vec(dS) = jS = (q)/(rho 4pi epsilon_(0) a^(2)) = (q)/(rho epsilon_(0))` But potential difference between the balls, `varphi_(+) - varphi_(-) = 2 (q)/(4pi epsilon_(0) a)` Hence, the sought resistanace, varphi `R = (varphi_(+) - varphi_(-))/(I) = (2q//4pi epsilon_(0) a)/(q//rho epsilon_(0)) = (rho)/(2pi a)` |
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| 40. |
Two conductors of aribitary shape are embedded into an infinite homogenous poorly conducting medium with respectivity `rho` and permittivity `epsilon`. Find the value of a product `RG` for this system where `R` is the resistance of the medium between the conductors, and `C` is the mutual capacitance of the wires in the presence of the medium. |
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Answer» Let us mentally impart charges `+q` and `-q` to the conductors. As the mediuim is poorly conducting, the surfaces of the conductors are equipotential and the field configuration is same as in the absense of the medium. Let us surround for example the positively chagred conductor, by a closed surface `S`, just containing the conductor, then, `R = (varphi)/(i) = (varphi)/(int vec(j). d vec(S)) , (varphi)/(int sigma E_(n) dS), as vec(j) uarr uarr vec(E)` and `C = (q)/(varphi) = (epsilon epsilon_(0) int E_(n) dS)/(varphi)` So, `RC = (epsilon epsilon_(0))/(sigma) = rho epsilon epsilon_(0)` |
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| 41. |
A charge `q` is distributed uniformly over the volume of a ball of radius `R`. Assuming the permittivity to be equal to unity, find : (a) the electrostatic self-energy of the ball, (b) the ratio of the energy `W_(1)` stored in the ball to the energy `W_(s)` pervadinting the surrounding space. |
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Answer» Electric fields inside and outside the sphere with the help of Gauss theorem : `E_(1) = (varphi r)/(4pi epsilon R^(2)) (r le R), E_(2) = (q)/(4pi epsilon_(0)) (1)/(r^(2)) (r gt R)` Sought self energy of the ball `U = W_(1) + W_(2)` `= int_(0)^(R) (epsilon_(0) E_(1)^(2))/(2) 4pi r^(2) dr + int_(R)^(oo) (epsilon_(0) E_(2)^(2))/(2) 4pi r^(2)n dr = (q^(2))/(8pi epsilon_(0) R) ((1)/(5) + 1)` Hence, `U = (3q^(2))/(4pi epsilon_(0) 5R)` and `(W_(1))/(W_(2)) = (1)/(5)` |
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| 42. |
A point charge `q = 3.0 m,uC` is located at the centre of a spherical layer of uniform isotropic dielectric with permittivity `epsilon = 3.0`. The inside radius of the layer is equal to `a = 250 mm`, the outside radius is `b = 500 mm`. Find the electrostatic energy inside the dielectric layer. |
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Answer» (a) By exprerssion `int (1)/(2) epsilon_(0) epsilon E^(2) dV = int (1)/(2) epsilon_(0) E_(2) dV = int (1)/(2) epsilon E^(2) 4pi r^(2) dr`, for a spherical layer. To find the electrostativ energy inside the dielectric layer, we have to intergrate the upper expression in the limit `[a,b]` `U = (1)/(2) epsilon_(0) epsilon int_(a)^(b) ((q)/(4pi epsilon_(0) epsilon r^(2)))^(2) 4pi r^(2) dr = (q^(2))/(8pi epsilon_(0) epsilon) [(1)/(a) - (1)/(b)] = 27 mJ` |
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| 43. |
A spherical shell is uniformly charged with the surface density `sigma`. Using the energy conservation law, find the magnitude of the electric force acting on a unit area of the shell. |
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Answer» Energy of the chagred sphere of radius `r`, from the equation `U = (1)/(2) q varphi = (1)/(2) q (q)/(4pi epsilon_(0) r) = (q^(2))/(8pi epsilon_(0) r)` If the radius of the shell changes by `dr` then work done is `4pi r^(2) F_(u) dr = -dU = q^(2)//8pi epsilon_(0) r^(2)` Thus sought force per unit area, `F_(u) = (q^(2))/(4pi r^(2) (8pi epsilon_(0) r^(2))) = ((4pi r^(2) sigma)^(2))/(4pi r^(2) xx 8pi epsilon_(0) r^(2)) = (sigma^(2))/(2 epsilon_(0))` |
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| 44. |
Let `B_(P)` and `B_(Q)` be the magnetic field produced by the wire `P` and `Q` which are placed symmetrically in a rectangular loop `ABCD` as shown in figure.Current in wire `P` is `I` directed inward and in `Q` is `2I` directed outwards.if `int_(A)^(B)vecB_(Q).vecdl=2mu_(0)` tesla meter,`int_(D)^(A)vecB_(P).vecdl=-2mu_(0)` tesla meter,&`int_(A)^(B)vecB_(P).vecdl=-mu_(0)` tesla meter tesla meter the value of `I` will be: A. `8 A`B. `4 A`C. `5 A`D. `6 A` |
| Answer» Correct Answer - D | |
| 45. |
A current of `2A` enters at the corner `d` of a square frame `abcd` of side `10 cm` and leaves at the opposite corner `b`.A magnetic field `B=0.1 T` exists in the space in direction perpendicular to the plane of the frame as shown in figure.Find the magnitude and direction of the magnetic forces on the four sides of the frame. |
| Answer» Correct Answer - A::B::C::D | |
| 46. |
A particle having mass `m` and charge `q` is released from the origin in a region in which electric field and magnetic field are ginen by `B=+B_(0)hatj` and `vecE=+E_(0)hati`. Find the speed of the particle as a function of its `X`-coordinate. |
| Answer» Correct Answer - B | |
| 47. |
The permeability of magnetic material is 0.9983. Name the type of magnetic materials it represents. |
| Answer» Magnetic material is paramagnetic in nature. | |
| 48. |
What amount of heat will be generated in a coil of resistance `R` due to a charge q passing through it if the current in the coil a. decreases down to zero uniformly during a time interval `t_0`? b. decrases down to zero having its value every `t_0` seconds? |
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Answer» (a) As current `i` is linear function of time, and at `t = 0` and `Delta r`, it equals `i_(0)` and zero respectively, it may be represented as, `i = i_(0) (1 - (t)/(Delta r))` Thus `q = int_(0)^(Delta t) i dt = int_(0)^(Delta t) i_(0) (1 - (t)/(Delta r)) dt = (i_(0) Delta t)/(2)` So, `i_(0) = (2q)/(Delta t)` Hence, `i = (2q)/(Delta t) (1 - (t)/(Delta r))` The heat generated. `H = int_(0)^(Delta t) i^(2) R dt = int_(0)^(Delta t) [(2q)/(Delta r) (1 - (t)/(Delta r))]^(2) R dt = (4q^(2) R)/(3 Delta r)` (b) Obviously the current through the coll is given by `i = i_(0) ((1)/(2))^(t//Delta t)` Then charge `q = int_(0)^(oo) idt = int_(0)^(oo) i_(0) 2^(-t//Delta t) dt = (i_(0) Delta t)/(In 2)` So, `i_(0) = (q In 2)/(Delta t)` And hence, heat generated in the circuit in the time interval `t[0, oo]`, `H = int_(0)^(oo) i^(2) R dt = int_(0)^(oo) [(q In 2)/(Delta t)2^(-t//Deltat)]^(2) R dr = (-q In 2)/(2 Delta t) R` |
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| 49. |
Fig illustrates a basic magnetization curve of iron (commerical purity grade). Using tis plot, draw th permeability `mu` as a function of the magnetic filed strength `H`. At what value of `H` is the permeability the greatest ? What is `mu_(max)` equal to ? |
| Answer» One has to draw the graph of `mu = (B)/(mu_(0) H)` versus `H` from the given. The `mu - H` graph starts out horizontally, and then rises steply at about `H = 0.04 k A//m` before falling again. It is easy to check that `mu_(max) = 10,000`. | |
| 50. |
A paramagnetic material is placed in a magnetic field. Consider the following statements : (A) If the magnetic field is increased, the magnetization is increased. (B) If the temperature is increased, the increased the magnetization is increased.A. Both `(a)` and `(b)` are trueB. `(a)` is true but `(b)` is falseC. `(b)` is true but `(a)` is falseD. Both `(a)` and `(b)` are false |
| Answer» Correct Answer - B | |