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

Options

A. 0.25 A
B. `0.50 A`
C. `0.75 A`
D. `1A`

Answer

C

Explanation

Given , `L = 60 cm, phi = 157 xx 10^(-6) Wb`
Magnetic induction inside the solenoid is
`B = (mu_(0) NI)/(L) "…"(i)`
Magnetic flux, `phi = BA`
`= (mu_(0) NI.A)/(L)` [ `:'` from Eq. (i)]
Magnetic moment `= NIA = (phiL)/(mu_(0))`
Substituting the given values, we get
`M = (157 xx 10^(-6) xx 0.6)/(4 xx 3.14 xx 10^(-7)) = 0.7A`