Free Objective Test 01 Practice Test - 11th and 12th
Question 1
Electrons move at right angle to a magnetic field of 1.5×10−2 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?
SOLUTION
Solution : C
r=mvqB⇒v(q/m).B=6×1071.7×1011×1.5×10−2=2.35×10−2m=2.35 cm
Question 2
An electron (mass = 9×10−31kg, charge =1.6×10−19 C ) whose kinetic energy is 7.2×10−20 J is moving in a circular orbit in a magnetic field of 9×10−5 weber/m2. The radius of the orbit is
SOLUTION
Solution : B
r=√2mKqB=√2×9×10−31×7.2×10−201.6×10−19×9×10−5=2.5 cm
Question 3
An electron and a proton enter a magnetic field perpendicularly. Both have same kinetic energy. Which of the following is true?
SOLUTION
Solution : B
By using r=√2mKqB; For both particles q → same, B → same, K → same
Hence r∝√m⇒rerp=√memp ∵mp>me so rp>re
Since radius of the path of proton is more, hence its trajectory is less curved.
Question 4
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?
SOLUTION
Solution : A
By using r=mvqB; v → same, B → same ⇒ r∝mq⇒rprα=mpmα×qαqp=mp4mp×2qpqp=12
Question 5
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?
SOLUTION
Solution : A
By using r=√2mKqB; r → same, B → same ⇒K∝q2m
Hence KαKp=(qαqp)2×mpmα=(2qpqp)2×mp4mp=1⇒Kα=Kp=1MeV.
Question 6
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
SOLUTION
Solution : D
12mv2=qV⇒v=√2qVm. Also F=qvB
⇒ F=qB√2qVm hence F∝√V which gives F′=√5F.
Question 7
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?
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=mvqB⇒r∝v
From given figure radius of the path described by particle B is more than that of A.
Hence vB>vA.
Question 8
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.
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.
Question 9
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
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
Question 10
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
SOLUTION
Solution : D
By applying Right hand thumb rule or Fleming's left hand rule, direction of force is found towards west.
Question 11
A charged particle of charge 4 mc enters a uniform magnetic field of induction →B=4→i+y→j+z→k tesla with a velocity →V=2→i+3→j−6→k. If the particle continues to move undeviated, then the strength of the magnetic filed induction (B) is
SOLUTION
Solution : A
→B||→V42=y3=z−6y=6,z=−12→B=4^i+6^j−12^k,B=√42+62+122B=√196=14
Question 12
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?
SOLUTION
Solution : C
F1F2=B i l1 cos45Bill sin 90∘F1F2=1
Question 13
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
SOLUTION
Solution : B
T=B I A n (0.8)
Sinθ=45θ=53∘
Question 14
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?
SOLUTION
Solution : A
T = BiAn
=0.02×50×10−3×12×10−4×200=24×10−5=2.4×10−4Nm
Question 15
A proton is moving in a field of induction [2⃗i+4⃗j]T with velocity [3×107⃗i]ms−1. The force experienced by it will be?
SOLUTION
Solution : A
⃗F=q(⃗V×⃗B)=1.6×10−19[3×107^i]×[2^i+4^j]=1.6×10−19×3×107×4^k⃗F=16.8×10−12^k