A solenoid of length `0*4m` and having `500` turns of wire carries a current of `3A`. A thin coil having 10 turns of wire and of radius `0*01m` carries a current of `0*4A`. Calculate the torque required to hold the coil in middle of the solenoid with its axis perpendicular to the axis of the solenoid.
`(mu_0=4pixx10^-7V-s//A-m)`

Explanation

For solenoid, `l_1=0*4m`, `N_1=500`,
`I_1=3A`
Magnetic field inside the solenoid is
`B=(mu_0N_1I_1)/(l_1)` acting along the axis of solenoid
For coil
`N_2=10`, `r=0*01m`, `I_2=0*4A`
Magnetic moment of the coil is
`M=N_2I_2A=N_2I_2pir^2` acting along the axis of coil, which is perpendicular to the axis of solenoid. So angle between `vecB` and `vecM` is `theta=90^@`.
Required torque`=MB sin theta`
`=N_2I_2pir^2xx(mu_0N_1I_1)/(l_1)sin90^@`
`=10xx0*4xxpixx(0*01)^2`
`xx(4pixx10^-7xx500xx3)/(0*4)xx1`
`=6pi^2xx10^-7=6xx9*87xx10^-7`
`=5*92x10^-6N-m`