InterviewSolution
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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 mass `m` charge `Q` and kinetic energy `T` eneters transverse unifrom magnetic fiedl of induction `vec(B)` After `s` the kinetic energy or the particle will beA. `T`B. ` 4T`C. `3 T`D. `2 T` |
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Answer» Correct Answer - A Magnetic force changes direction of velocity, not its magnitude hence kinetic energy remains same. |
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| 2. |
A charged particle is fired at an angle `theta` to a uniform magnetic field directed along the x-axis. During its motionalong a helical path, the particle willA. (i), (ii)B. (ii), (iii)C. (i), (iii)D. (i), (iv) |
| Answer» Correct Answer - D | |
| 3. |
Two particles `A and B` of masses ` m_(A) and m_(B)` respectively and having the same charge are moving in a plane . The speeds of the particles are ` v_(A) and v_(B)` respectively and the trajectories are as shown in the figure. Then A. `m_(A) v_(A) lt m_(B) v_(B)`B. `m_(A) v_(A) gt m_(B) v_(B)`C. `m_(A) v_(B) lt m_(A) v_(B)`D. `m_(A) = m_(B), v_(A) = v_(B)` |
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Answer» Correct Answer - B `R_(A) gt R_(B)` `(m_(A)v_(A))/(Bq) gt (m_(B)v_(B))/(Bq) rArr m_(A)v_(A) gt m_(B)v_(B)` |
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| 4. |
A current carrying loop is free to turn in a uniform magnetic field.The loop will then come into equilibrium when its plane is inclined atA. `0^(@)` to the direction of the fieldB. `45^(@)` to the directed of the fieldC. `90^(@)` to the directed of the fieldD. `135^(@)` to the directed of the field |
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Answer» Correct Answer - C `U =-vec(M) . vec(B) = -MB cos theta` Equilibrium will be stable if `U` is minimum. It will happen when `theta =0^(@), U_(min) = -MB` |
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| 5. |
A particle of charge `-16xx10^(-18) coulomb` moving with velocity `10 ms^(-1)` along the ` x- axis `, and an electric field of magnitude `(10^(4))//(m)` is along the negative ` z- ais`. If the charged particle continues moving along the ` x`- axis , the magnitude of `B` isA. `10^(-3) Wb//m^(2)`B. `10^(3) Wb//m^(2)`C. `10^(5) Wb//m^(2)`D. `10^(16) Wb//m^(2)` |
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Answer» Correct Answer - B Since particle goes in a straight line `F_(e) = F_(m) rArr eE = Bev` `B = (E)/(v) = (10^(4))/(10) = 10^(3) Wb//m^(2)` |
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| 6. |
A particle having mass m and charge `q` is released from the origin in a region in which ele field and magnetic field are given by `B=-B_0hatj`and `E=E_0hatk`. Find the y- component of the velocity and the speed of the particle as a function of it z-coordinate.A. `sqrt((2qE_(0)z)/(m))`B. `sqrt((qE_(0)z)/(m))`C. `sqrt((qE_(0)z)/(2m))`D. `sqrt((4qE_(0)z)/(m))` |
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Answer» Correct Answer - A The work is done by electric force. Work done `=` change in kinetic energy `W_(1 rarr 2) = Delta K` `qE_(0)z = (1)/(2)mv^(2) - 0` `v = sqrt((2qE_(0)z)/(m))` |
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| 7. |
A proton, a deuteron and an `alpha`- particle having the same kinetic energy are moving in circular trajectors in a constant magnetic field. If `r_p, r_d` and `r_(alpha)` denote respectively the radii of the trajectories of these particles thenA. `r_(alpha) = r_(p) lt r_(d)`B. `r_(alpha) gt r_(d) gt r_(p)`C. `r_(alpha) = r_(d) lt r_(p)`D. `r_(p) = r_(d) = r_(alpha)` |
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Answer» Correct Answer - A `r = sqrt(2mK)/(Bq), r_(p) prop (sqrt(m))/(e), r_(d) prop (sqrt(2m))/(e), r_(alpha) prop (sqrt(4m))/(2e)` `r_(p) = r_(alpha) lt r_(d)` |
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| 8. |
A circular loop of radius a, carrying a current I, is placed in a tow dimensional magnetic field. The centre of the loop coincides with the centre of the filed The strenght of the magnetic field at the pariphery of the loop is B. find the magnetic force on the wire. A. `2pi B i a`B. `4pi B i a`C. `pi B i a`D. `B i a` |
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Answer» Correct Answer - A Here direction of magnetic field is different at every point i.e. magnetic field is non-uniform. Consider a small length `dl`. `dF = B i d l` `F = Bi int dl = Bi. 2pi a = 2pi Bia` |
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| 9. |
The current flows in the wire from `A` to `D` clockwise direction. The net force on `ABCD` A. `(sqrt(17))/(2)BiR`B. `(sqrt(15))/(2)BiR`C. `(sqrt(13))/(2)BiR`D. `B i R` |
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Answer» Correct Answer - A `F_(AB) = B i R`, towards left `F_(BC) = Bi.2R = 2BiR`, upwards `F_(CD) = (BiR)/(2)`, towards right `F_(n et) = sqrt((BiR - (BiR^(2))/(2))^(2) + (2BiR)^(2)) = (sqrt(17) BiR)/(2)` |
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| 10. |
A charged particle moves in a gravity-free space without change in velocity. Which of the following is/are possible? (i) `E = 0, B = 0` (ii) `E = 0, B != 0` (iii) `E != 0, B = 0` (iv) `E != 0, B != 0`A. (i), (ii)B. (ii), (iv)C. (ii), (iii)D. (i), (ii), (iv) |
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Answer» Correct Answer - D Since velocity is constant, net force `= 0` (i) `E = 0, B = 0` i.e. `F_(e) = 0, F_(m) = 0` O.K. (ii) `E = 0, B != 0` i.e. `F_(e) = 0`, It may be that velocity is parallel or anti-parallel to magnetic field and hence `F_(m) = 0` O.K. (iii) `E != 0 , B = 0`, The electric will change magnitude or magnitude and direction or velocity, Not O.K. (iv) `E != 0, B != 0`, It may be that electric field balances magnetic force. O.K. |
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| 11. |
The figure shows a circular coil and a long straight wire `AB` placed close to each other, the wire being parallel to a diameter of coil. The arrows show directions of currents. The direction of magnetic force acting on `AB` is A. out of the pageB. into the pageC. towards rightD. towards left |
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Answer» Correct Answer - D Magnetic field due to ring is in upward direction `vec(F)_(m) = i(- l hat(j) xx B hat(k)) = - i l B hat(i)` |
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| 12. |
A thin wire of length `L` is connected to two adjacent fixed points and carries a current `I` in the clockwise direction , as shown in the figure. When the system is put in a uniform magnetic field of strength `B` going into the plane of the paper , the wire takes the shape of a circle . The tension in the wire is A. `IBL`B. `(IBL)/(pi)`C. `(IBL)/(2pi)`D. `(IBL)/(4pi)` |
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Answer» Correct Answer - C `L = 2pi r rArr r = (L)/(2pi)` `T = Bir = (BIL)/(2pi)` |
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| 13. |
A closed loop lying in the `xy` plane carries a current. If a uniform magnetic field `B` is present in the region, the force acting on the will be zero if `B` is inA. the `x`-dierctionB. the `y`-dierctionC. the `z`-dierctionD. ant of the above direction |
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Answer» Correct Answer - D If closed loop carrying a single current is placed in uniform magnetic field, net magnetic force on loop is zero. |
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| 14. |
if a charged particle projected in a gravity free room deflects,A. (i), (ii)B. (ii), (iv)C. (ii), (iii)D. (iii) , (iv) |
| Answer» Correct Answer - D | |
| 15. |
A magnetic field ` vec(B) = B_(0)hat(j)` , exists in the region `a ltxlt2a` , and `vec(B) = -B_(0) hat(j)` , in the region `2a lt xlt 3a`, where `B_(0)` is a positive constant . A positive point charge moving with a velocity `vec(v) = v_(0) hat (i)` , where `v_(0)` is a positive constant , enters the magnetic field at `x= a` . The trajectory of the charge in this region can be like A. B. C. D. |
| Answer» Correct Answer - A | |
| 16. |
A very long straight wire carries a current I. At the instant when a charge `+Q` at point `P` has velocity `vecV`, as shown, the force on the charge is A. opposite to `OX`B. along `OX`C. opposite `OY`D. along `OY` |
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Answer» Correct Answer - D `vec(F)_(m) = q(v hat(i) xx B hat(k)) = q v B hat(j)` |
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| 17. |
Four wires each of length `2.0` meters area bent into four loops `P,Q,R` and `S` and then suspended into uniform magnetic field. Same current is passed in each loop. Which statement is correct? A. `P`B. `Q`C. `R`D. `S` |
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Answer» Correct Answer - D For given length, circle has maximum area. `tau prop M prop A` `tau prop A` |
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| 18. |
A coil in the shape of an equilateral triangle of side `l` is suspended between the pole pieces of permanent magnet. Such that `vecB` is in plane of the coil. If due to a current I in the triangle, a torque `tau` acts on it, the side l of the triangel is:A. `(2)/(sqrt(3))((tau)/(Bi))^((1)/(2))`B. `(2)/(3)((tau)/(Bi))`C. `2((tau)/(sqrt(3) Bi))^((1)/(2))`D. `(1)/(sqrt(3)) (tau)/(Bi)` |
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Answer» Correct Answer - C `tau = MB sin 90^(@) = MB` `tau = iAB = i((sqrt(3) l^(2))/(4))B` `l^(2) = (4 tau)/(sqrt(3)iB) rArr l = 2((tau)/(sqrt(3) Bi))^((1)/(2))` |
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| 19. |
Two insulated rings, one of a slighlty smaller diameter than the other are suspended along their common diameter as shown. Initially the planes of the rings are mutually perpendicular. When a steady current is set up each of them A. The two rings rotate into a common planeB. The inner rings oscillates about its initial positionC. The inner ring stays stationary while the outer one moves into the plane of the inner ringD. The outer ring stays stationary while the inner one moves into the plane of the outer ring |
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Answer» Correct Answer - A The torque will be exerted on rings. |
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| 20. |
A charged particle begins to move in a magnetic field, initially parallel to the field. The direction of the field now begins to change, with its magnitude remaining constantA. The magnitude of the force acting on the particle will remain constantB. The magnitude of the force acting on the particle will changeC. The particle will always move parallel to the fieldD. The speed of the particle will change |
| Answer» Correct Answer - B | |
| 21. |
Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will A. remain stationaryB. move towards the wireC. move away from the wireD. rotate about the wire |
| Answer» Correct Answer - B | |
| 22. |
At a specific instant emission of radioactive compound is deflected in a magnetic field . The compound can emit (i) electron (ii)protons(iii)`He^(2+ )`(iv) neutrons The emission at instant can beA. (i), (ii), (iii)B. allC. (iv) onlyD. (ii), (iii) |
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Answer» Correct Answer - A Only charged particles are deflected in magnetic field. |
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| 23. |
A stream of protons and `alpha`-particle of equal momenta enter a unifom magnetic field perpendicularly. The radii of their orbits are in the ratioA. (1) `1`B. `2`C. `0.5`D. `4` |
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Answer» Correct Answer - B `R = (p)/(Bq), (R_(p))/(R_(alpha)) = (2e)/(e) = 2` |
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| 24. |
In the previous question, (i) the force on `T` is proportional to `i` (ii) the force `T` is proportional to `i^(2)` (iii) if direction of `I` is reversed in the circuit be reversing `E`, the force on `T` will remain in the same directionA. (i), (ii)B. (ii), (iv)C. (ii), (iii)D. (i), (iv) |
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Answer» Correct Answer - B `B prop i, F_(m) prop i^(2)` `vec(F)_(m) = i(-l hat(j) xx -B hat(k)) = i l B hat(i)` |
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| 25. |
A particle of charge `+q` and mass `m` moving under the influence of as uniform electric field `Ehati` and uniform magnetic field `B hatk` follows trajectory form `P` to `Q` as shown in figure. The velocities at P and Q re `vhati and -2hatj`. Which of the following statement(s) is/are correct? `a. E=3/4[(mv^2)/(qa)]` b. Rate of work done by the electrif field at P is `3/4[(mv^3)/a]` c. Rate of work done by the electric field at `P` is zero. d. Rate of work done by both the field at `Q` is zero.A. (i), (ii)B. (ii), (iv)C. (i), (ii), (iv)D. (i), (iv) |
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Answer» Correct Answer - C (i) Work is done only by electric force. By work-energy theorem between `P` and `Q` `Delta W = K_(2) - K_(1) = K_(Q) - K_(P)` `qE xx 2a = (1)/(2)m(2v^(2)) - (1)/(2)mv^(2) = (3)/(2)mv^(2)` `E = (3mv^(2))/(4qa)`, (i) is O.K. (ii) `P = vec(F)_(e). vec(v) = qvec(E).vec(v) = q(Ehat(i)).(v hat(i)) = qEv` (iv) Work done by magnetic force is always force is always zero At `Q`, `F_(e)_|_^(ar)v, P = 0` (iv) is O. K. |
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