Question

A circular coil of 20 turns and 10 cm radius 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 A, cross sectional area is 10-5 m2 and coil is made up of copper wire having free electron density about 1029 m-3, then the average force on each electron in the coil due to magnetic field is

• A. 2.5 × 10-25 N
• B. 5 × 10-25 N
• C. 4 × 10-25 N
• D. 3 × 10-25 N

Answer 1

Answer: (B)

5 × 10-25 N

Answer 2

Force on each electron

$\left[ \begin{array} { l } \because \mathrm { I } = \mathrm { neAv } \mathrm { ve } _ { \mathrm { d } } \\ \therefore \mathrm { ev } _ { \mathrm { d } } = \frac { \mathrm { I } } { \mathrm { nA } } \end{array} \right]$

Here I = 5A, B $\mathrm { A } = 10 ^ { - 5 } \mathrm {~m} ^ { 2 }$

So,