The range flux near the axis and inside the air core solenoid of length 60 cm carrying current 'I' is `1.7 xx 10^(-6)Wb`. Its magnetic moment will be (cross sectional area of a solenoid is very small as comapred to its length,`mu_0=4pi xx 10^(-7)SI unit)`

Options

A. `0.25 Am^(2)`
B. `0.50 Am^(2)`
C. `0.75 Am^(2)`
D. `1 Am^(2)`

Answer

C

Explanation

Magnetic induction inside the solenoid `B=(mu_0NI)/L` and magnetic flux`phi=BA therefore phi=(mu_0NI)/Lx A` therefore Magnetic moment=`NIA=(phiL)/(mu_0)=(1.57 xx 10^(-6)xx 0.6)/(4 xx 3.14xx 10^(-7))=(0.6 xx 10^(1))/(4 xx 2)=3/4 therefore M=0.75 Am^(2)`.