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 5.0 A, what is the(a) total torque on the coil,(b) the total force on the coil,(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area $10^{- 5} m^{2}$ , and the free electron density in copper is given to be about $10^{29}$ $m^{- 3} .$ )

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 5.0 A, what is the(a) total torque on the coil,(b) the total force on the coil,(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area $10^{- 5} m^{2}$ , and the free electron density in copper is given to be about $10^{29}$ $m^{- 3} .$ )

Toppr Admin · Accepted Answer

-a-The total torque on coil is zero because the field is uniform-b-The-total force on coil is zero as field is uniform-c-Force-F-Be-xD7-Vd-Vd-drift velocity-INeA-F-5-xD7-10-x2212-25N on each electron-

(a)The total torque on coil is zero because the field is uniform.(b)The total force on coil is zero as field is uniform.(c)Force, F=Be×Vd Vd=drift velocity=INeA F=5×10−25N on each electron.

(a)The total torque on coil is zero because the field is uniform.
(b)The total force on coil is zero as field is uniform.
(c)Force, $F = B_{e} \times V_{d}$

$V_{d} = drift velocity = \frac{I}{N e A}$

$F = 5 \times 10^{- 25} N$ on each electron.