Discussion Forum



1)

The magnetic flux near the axis and inside  the air-core solenoid of length 60 cm carrying current 'l' is 1.57 x 10-6 Wb. its magnetic moment will be (cross-sectional area  of a solenoid is very small as compared to its length, $\mu_{0}=4\pi \times 10^{-7}$ SI unit)


A) 0.25 A

B) 0.50 A

C) 0.75 A

D) 1 A

Answer:

Option C

Explanation:

Given, L= 60cm, $\phi$= 1.57 x 10-6Wb

 Magnetic induction inside the solenoid is 

$B=\frac{\mu_{0}Nl}{L}$  .......(i)

 Magnetic flux, $\phi$  =BA

     $=\frac{\mu_{0}Nl.A}{L}$     [ $\because$  from Eq.(i)]

 Magnetic  moment = NlA $ =\frac{\phi L}{\mu_{0}}$

 Substituting the given values, we get

$M=\frac{1.57 \times 10^{-6}\times0.6} {4 \times 3.14 \times 10^{-7}}=0.75 A$