Question

A circular coil of $ 20\,turns\, $ , and radius $ 10\,cm $ is placed in a uniform magnetic field of $ 0.10\,T $ normal to the plane of the coil. If the current in the coil is $ 5A $ , (a) what is the total torque on the coil?, (b) total force on the coil (c) average force on each electron in the coil due to the magnetic field?
(The coil is made up of copper wire of cross sectional area $ {10^{ - 5}}\,{m^2} $ , and the free electron density in copper is given to be about $ {10^{29}}\,{m^{ - 3}} $ )

Answer 1

Hint: In order to solve this question, we are going to take the values of the magnetic field, current and the radius of the coil to find the torque on the coil, the, after that we find the net force on the coil, then the average force on the each electron in the coil due to magnetic field is found.
Torque is given by
 $ \tau = NBA\sin \theta $
 $ N $ is the number of turns, $ B $ is the magnetic field and $ I $ is the current in the coil.
If $ {v_d} $ is drift velocity of electron, then,
 $ F = qv \times B \\
  F = e{v_d}B\sin {90^ \circ } \\ $

Complete step by step solution:
 It is given that the number of turns of the coil, $ N = 20 $
And the radius of the coil, $ r = 10cm = 10 \times {10^{ - 2}}m $
The magnetic field is given, $ B = 0.10T $
Current carried by the coil is, $ I = 5A $
Angle between field and normal to the coil
Area of the coil,
 $ A = \pi {r^2} \\
  A = \pi \times {\left( {10 \times {{10}^{ - 2}}} \right)^2} \\
  A = \pi \times {10^{ - 2}}{m^2} \\ $
(a) Torque is given by
 $ \tau = NBA\sin \theta $
Putting the values, in this equation, we get
 $ \tau = 20 \times 5 \times 0.10 \times \pi \times {10^{ - 2}}\sin {0^ \circ } \\
  \tau = 20 \times 5 \times 0.10 \times \pi \times {10^{ - 2}} \times 0 \\ $
(b) Net force on a planar current loop in a magnetic field is always zero, as net force due to a couple of forces is zero.
(c) If $ {v_d} $ is drift velocity of electron , then,
 $ F = qv \times B \\
  F = e{v_d}B\sin {90^ \circ } \\ $
 $ The{\text{ }}force{\text{ }}on{\text{ }}one{\text{ }}electron = Be{v_d} = Be\dfrac{l}{{neA}} = \dfrac{{B\operatorname{l} }}{{nA}} $
Putting the values, we get,
 $ The{\text{ }}force{\text{ }}on{\text{ }}one{\text{ }}electron = \dfrac{{0.10 \times 5.0}}{{{{10}^{29}} \times {{10}^{ - 5}}}} = 5 \times {10^{ - 25}}N $

Note:
It is important to note that as the angle between the velocity and magnetic field is zero degree, this makes the torque of the coil zero also, the net magnetic force is zero as net force due to couple force is zero. The force on one electron is found by dividing net force by the number of electrons.