A solenoid of cross-sectional area `2xx 10^(-4)m^(2)` and 900 turns has `0.6A m^(2)` magnetic moment. Then the current flowing through it is

Options

A. 2.4A
B. 2.34mA
C. 3.33A
D. 3.33mA

Answer

C

Explanation

Here, N=900 turns
`A=2xx10^(-4)m^(2), m_(s)=0.6Am^(2)`
The magnetic moment of solenoid `m_(s)=NIA`
The current flowing through the solenoid is
`I=(m_(s))/(NA)=(0.6)/(900xx2xx10^(-4)=3.33A`