A small coil of radius 0.002 m is placed on the axis of a magnet of magnetic moment `10^(5)` `JT^(-1)` and length 0.1 m at a distance of 0.15 m from the centre of the magnet. The plane of the coil is perpendicular to the axis of the magnet. Find the force on the when a current of 2.0 A is passed through it.

Answer

`04.40`

Explanation

`a= 0.002m`
`I= 2A`
`2l = 0.1m`
` r= 0.15m`
` M = 10^5 JT^(-1)`
`M = 10^5 JT^(-1)`
Now, pole strength of magnet,
`m=M/(2l)= 10^5/(0.1) =10^6 ` Am
Now, Magnetic field at a distance x from the centre of a current carrying coil of radius .a.,
`B = (mu_0)/(4pi) .(2pi Ia^2)/(x^3) (a lt lt x)`
If `B_1 and B_2` are magnetic fields due to the north and south poles of the magnet, then force on the magnet due to the coil,
`F_("MC") = mB_(2) - mB_(1)`
Force on the coil due to magnet
`F_(CM) = -F_(MC) = mB_1 – mB_2`
`=m[mu_0/(4pi).(2piIa^2)/((r-l)^3) -(mu_0)/(4pi).(2piIa^2)/((r+l)^3)]`
`=mu_0/(4pi).2pila^2 m [1/((r-l)^(3)) - (1)/((r-l)^3)]`
`=10^(-7)xx2pi xx 2 xx (0.002 )^2 xx10^6`
`xx[1/((0.15 - 0.05)^3)-1/((0.15 + 0.05)^3)]`
`= 5.024 xx 10^(-6) (1000 -125)`
`:. F_(CM) = 4.4 xx10^(-3) N`