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A small bar magnet has a dipole moment 1.2 A m^2. The magnitude of magnetic field at a distance of 0.1 m on its axis will be

Answer»

`1.2xx10^(-4)T`
`2.4 XX 10^(-4) T`
`2.4 xx 10^(4) T`
`1.2xx10^(4)T`

Solution :Along AXIAL line of bar MAGNET `B = (mu_0)/(4pi) * (2m)/(r^3) = (10^(-7) xx 2xx 1.2)/((0.1)^3) = 2.4 xx 10^(-4) T`


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