Physics Aptitude Test for Schools 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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
Answer: d
Clarification: 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
Answer: c
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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