Physics Aptitude Test for Schools

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
View Answer

Answer: c
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
View Answer

Answer: b
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
View Answer

Answer: a
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.

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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
View Answer

Answer: d
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
View Answer

Answer: c
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

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

Answer: b
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
View Answer

Answer: b
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
View Answer

Answer: a
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
View Answer

Answer: c
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
View Answer

Answer: b
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

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