A compass needle whose magnetic moment is `60Am^2` pointing geograhic north at a certain place, where the horizontal component of earth's magnetic field is `40muWbm^-2` experiences a torque `1*2xx10^-3Nm`. What is the declination of the place?

Answer

`30^@`

Explanation

Here, `I=3*3A`, `B=0*33G=0*33xx10^-4T`, `delta=0^@`. Horizontal component of earth's magnetic field `H=Bcos delta=0*33xx10^-4cos0^@=0*33xx10^-4T`. The neutral points will be parallel and above the cable. Let r be the distance of neutral point from the cable. Then
`H=(mu_0)/(4pi)(2I)/(r)` or `r=(mu_0)/(4pi)xx(2I)/(H)`
`r=10^-7xx(2xx3*3)/(0*33xx10^-4)=2xx10^-2m=2cm`
Thus the set of neutral points will lie parallel to cable, and above the cable at distance of `2cm` from cable.