# Free Objective Test 01 Practice Test - 11th and 12th

Electrons move at right angle to a magnetic field of 1.5×102 Tesla with a speed of 6×107 m/s. If the specific charge of the electron is 1.7×1011 C/kg, then the radius of the circular path will be?

A. 2.9 cm
B. 3.9 cm
C. 2.35 cm
D. 3 cm

#### SOLUTION

Solution : C

r=mvqBv(q/m).B=6×1071.7×1011×1.5×102=2.35×102m=2.35 cm

An electron (mass = 9×1031kg, charge =1.6×1019 C ) whose kinetic energy is 7.2×1020 J is moving in a circular orbit in a magnetic field of  9×105 weber/m2. The radius of the orbit is

A. 1.25 cm
B. 2.5 cm
C. 12.5 cm
D. 25.0 cm

#### SOLUTION

Solution : B

r=2mKqB=2×9×1031×7.2×10201.6×1019×9×105=2.5 cm

An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic energy. Which of the following is true?

A. Trajectory of electron is less curved
B. Trajectory of proton is less curved
C. Both trajectories are equally curved
D. Both move on straight line path

#### SOLUTION

Solution : B

By using r=2mKqB;   For both particles q  same, B  same, K   same
Hence rmrerp=memp   mp>me so rp>re
Since radius of the path of proton is more, hence its trajectory is less curved.

A proton and an α particle enter a uniform magnetic field with same velocity, then ratio of the radii of path describe by them will be?

A. 1 : 2
B. 1 : 1
C. 2 : 1
D. None of these

#### SOLUTION

Solution : A

By using r=mvqB; v   same, B   same  rmqrprα=mpmα×qαqp=mp4mp×2qpqp=12

A proton of mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1MeV. What should be the energy of α -particle (mass = 4 m and charge = +2e), so that it can revolve in the path of same radius?

A. 1 MeV​
B. 4 MeV​
C. 2 MeV​
D. 0.5 MeV​

#### SOLUTION

Solution : A

By using r=2mKqB; r  same, B  same Kq2m
Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1Kα=Kp=1MeV.

A charge particle having charge q is accelerated through a potential difference V and then it enters a perpendicular magnetic field in which it experiences a force F. If V is increased to 5V, the particle will experience a force

A. F
B. 5F
C. F5
D. 5F

#### SOLUTION

Solution : D

12mv2=qVv=2qVm.  Also F=qvB
F=qB2qVm hence FV which gives F=5F.

The magnetic field is perpendicularly into the plane of the paper and a few charged particles are projected in it. Which of the following statement is true?

A. A represents proton and B and electron
B. Both A and B represent protons but velocity of A is more than that of B
C. Both A and B represents protons but velocity of B is more than that of A
D. Both A and B represent electrons, but velocity of B is more than that of A

#### SOLUTION

Solution : C

Both particles are deflecting in same direction so they must be of same sign.(i.e., both A and B represents protons)
By using r=mvqBrv
From given figure radius of the path described by particle B is more than that of A.
Hence vB>vA.

A metallic block carrying current i is subjected to a uniform magnetic induction B as shown in the figure. The moving charges experience a force F given by ……. which results in the lowering of the potential of the face …….? Assume the speed of the carriers to be v.

A. eVB^k, ABCD

B. eVB^k, ABCD
C. eVB^k, ABCD

D. eVB^k, EFGH

#### SOLUTION

Solution : C

As the block is of metal, the charge carriers are electrons; so for current along positive x-axis, the electrons are moving along negative x-axis, i.e. v=vi
and as the magnetic field is along the y-axis, i.e. B=B^j
so F=q(v×B) for this case yield F=(e)[v^i×B^j]
i.e., F=evB^k      [As ^i×^j=^k]
As force on electrons is towards the face ABCD, the electrons will accumulate on it an hence it will acquire lower potential.

An ionised gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +ve x-axis and a magnetic field along the +z direction then

A. Positive ions deflect towards +y direction and negative ions towards – y direction
B. All ions deflect towards + y direction
C. All ions deflect towards – y direction
D. Positive ions deflect towards – y direction and negative ions towards + y direction.

#### SOLUTION

Solution : C

As the electric field is switched on, positive ion will start to move along positive x-direction and negative ion along negative x-direction. Current associated with motion of both types of ions is along positive x-direction. According to Flemings left hand rule or right hand thumb rule, force on both types of ions will be along negative y-direction

A vertical wire carrying a current in the upward direction is placed in a horizontal magnetic field directed towards north. The wire will experience a force directed towards

A. North
B. South
C. East
D. West

#### SOLUTION

Solution : D

By applying Right hand thumb rule or Fleming's left hand rule, direction of force is found towards west.

A charged particle of charge 4 mc enters a uniform magnetic field of induction B=4i+yj+zk tesla with a velocity V=2i+3j6k. If the particle continues to move undeviated, then the strength of the magnetic filed induction (B) is___ tesla.

A. 14
B. 2
C. 3.5
D. 28

#### SOLUTION

Solution : A

B||V42=y3=z6y=6,z=12B=4^i+6^j12^k,B=42+62+122B=196=14

A wire abc is carrying current I. It is bent as shown in the figure and is placed in a uniform magnetic field B. Length ab = l and abc=45. What is the ratio of force on bc to that on ab?

A. 2
B. 12
C. 1
D. 32

#### SOLUTION

Solution : C

F1F2=B i l1 cos45Bill sin 90F1F2=1

A coil carries a current and experiences a torque due to a magnetic filed. The value of the torque is 80% of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is

A. 30
B. 53
C. 45
D. 37

#### SOLUTION

Solution : B

T=B I A n (0.8)
Sinθ=45θ=53

A rectangular coil 6 cm long and 2 cm wide is placed in a magnetic field of 0.02 T. If the loop contains 200 turns and carries a current of 50 mA, the torque on the coil when its area vector is perpendicular to the field will be?

A. 2.4×104Nm
B. 2.4 Nm
C. 0.24 Nm
D. 0.024 Nm

#### SOLUTION

Solution : A

T = BiAn
=0.02×50×103×12×104×200=24×105=2.4×104Nm

A proton is moving in a field of induction [2i+4j]T with velocity [3×107i]ms1. The force experienced by it will be?

A. 16.8×1012(+^k)N
B. 6×107(^k)N
C. 12×107(+^k)N
D. 0

#### SOLUTION

Solution : A

F=q(V×B)=1.6×1019[3×107^i]×[2^i+4^j]=1.6×1019×3×107×4^kF=16.8×1012^k