# Physics Questions and Answers – Motion in Combined Electric and Magnetic Field

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This set of Physics Aptitude Test for Schools focuses on “Motion in Combined Electric and Magnetic Field”.

1. Calculate the speed of an electron if it travels in a circular path of radius 50 cm in a magnetic field of 5 × 10-3 T.
a) 440 × 107 m/s
b) 4 × 107 m/s
c) 44 × 107 m/s
d) 0.4 × 107 m/s

Explanation: Magnetic force on the electron = Centripetal force on the electron.
evB = $$\frac {mv^2}{r}$$,
v = $$\frac {eBr}{m}$$
v = $$\frac {(1.6 \, \times \, 10^{-19} \, \times \, 5 \, \times \, 10^{-3} \, \times \, 50 \, \times \, 10^{-2})}{(9.1 \, \times \, 10^{-31})}$$
v = 43.95 × 107 m/s ≈ 44 x 107 m/s.

2. A chamber is maintained at a uniform magnetic field of 5 × 10-3 T. An electron with a speed of 5 × 107 ms-1 enters the chamber in a direction normal to the field. Calculate the radius of the path.
a) 15.7 cm
b) 5.7 cm
c) 25.7 cm
d) 75.7 cm

Explanation: Radius → r = $$\frac {mv}{eB}$$
r = $$\frac {(9.1 \, \times \, 10^{-31} \, \times \, 5 \, \times \, 10^7)}{(1.6 \, \times \, 10^{-19} \, \times \, 5 \, \times \, 10^{-3})}$$
r = 5.7 cm
Therefore, the radius of the path is 5.7 cm.

3. Which of the following particles cannot be accelerated using a cyclotron?
a) Neutron
b) α-particle
c) Deuteron
d) Proton

Explanation: Neutrons, being electrically neutral cannot be accelerated in a cyclotron. The cyclotron is a heavy particle accelerator used to accelerate charged particles like protons, deuterons and α-particles to high velocities.

4. If an electron entering at a magnetic field of 2 × 10-2 T has a velocity of 3 × 107 ms-1 and describes a circle of radius 8 × 10-3 m, then find the value of $$\frac {e}{m}$$ of the electron.
a) 1.567 × 1011 C/Kg
b) 190 × 1011 C/Kg
c) 18 × 1011 C/Kg
d) 1.875 × 1011 C/Kg

Explanation: Radius, r=$$\frac {mv}{eB}$$.
$$\frac {e}{m} = \frac {v}{rB}$$
$$\frac {e}{m} = \frac {3 \, \times \, 10^7}{(8 \, \times 10^{-3} \, \times \, 2 \, \times \, 10^{-2})}$$
$$\frac {e}{m}$$ = 1.875 × 1011 C/Kg.

5. A cyclotron has an oscillatory frequency of 12 MHz and a dee radius of 50 cm. Calculate the magnetic field required to accelerate deuterons of mass 3.3 × 10-27 Kg and charge 1.6 × 10-19 C.
a) 25.6 T
b) 75.5 T
c) 1.56 T
d) 7.56 T

Explanation: Cyclotron frequency = $$\frac {qB}{2 \pi m}$$.
B = $$\frac {(2 \pi mf_c)}{(q)}$$
B = $$\frac {(2 \, \times 3.142 \, \times 3.3 \, \times \, 10^{-27} \, \times \, 12 \, \times \, 10^6)}{(1.6 \, \times \, 10^{-19})}$$
B = 1.56 T

6. Electrons are accelerated in a cyclotron.
a) True
b) False

Explanation: Electrons cannot be accelerated in a cyclotron. A large increase in their energy increases their velocity to a very large extent. This throws the electrons out of step with the oscillating field. A cyclotron can only accelerate particles such as protons, deuterons, and alpha – particles.

7. The frequency of revolution of a charged particle in a cyclotron does not depend on ‘X’. Identify X.
a) Magnetic field
b) Speed of the particle
c) Mass of the particle
d) Charge on the particle

Explanation: The radius of the circular path of the charged particle increases in direct proportion to its speed. Consequently, both its time-period and frequency of revolution are independent of its speed. So, the ‘X’ is the speed of the particle.

8. Alpha particles of mass 6.68 × 10-27 Kg and charge 3.2 × 10-19 C is accelerated in a cyclotron in which a magnetic field of 1.25 T is applied perpendicular to the dees. How rapidly should the electric field between the dees be reversed?
a) 5.25 × 10-8 s
b) 955.25 × 10-8 s
c) 55.25 × 10-8 s
d) 575.25 × 10-8 s

Explanation: Time period, t = $$\frac {\pi m}{qB}$$
t = $$\frac {(3.14 \, \times \, 6.68 \, \times \, 10^{-27})}{(3.2 \, \times \, 10^{-19} \, \times 1.25)}$$
t = 5.25 × 10-8 s.

9. What should be the minimum magnitude of the magnetic field that must be produced at the equator of earth so that a proton may go round the earth with a speed of 1 × 107 ms-1? Earth’s radius is 6.4 × 106 m.
a) 16.63 × 10-8 T
b) 91.63 × 10-8 T
c) 1.63 × 10-8 T
d) 761.63 × 10-8 T

Explanation:B = $$\frac {mv}{qr}$$.
B = $$\frac {(1.67 \, \times \, 10^{-27} \, \times \, 10^7)}{(1.6 \, \times \, 10^{-19} \, \times \, 6.4 \, \times \, 10^6)}$$
B = 1.63 × 10-8 T
Therefore, the minimum magnitude of magnetic field should be 1.63 × 10-8 T.

10. In a cyclotron, a magnetic induction of 1.4 T is used to accelerate protons. How rapidly should the electric field between the dees be reversed?
a) 5.2 × 10-8 s
b) 2.34 × 10-8 s
c) 792.25 × 10-8 s
d) 46.25 × 10-8 s

Explanation: Time period, t = $$\frac {\pi m}{qB}$$
t = $$\frac {(3.14 \, \times 1.67 \, \times \, 10^{-27})}{(1.6 \, \times \, 10^{-19} \, \times \, 1.4)}$$
t = 2.34 × 10-8 s

Sanfoundry Global Education & Learning Series – Physics – Class 12.

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