Question

A bar magnet is 10cm long.Its magnetic moment is 0.4 Am^2.It is placed in magnetic meridian with its north pole towards north. If the neutral point is obtained at distance if 10cm from the center of the magnet find the horizontal component if earth's magnetic field.

Answer 1

Answer:-

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• Given:-

$m= 0.4 \: Am^2 \\ r = 10cm \\ 2l = 10cm$

• Solution:-

When North Pole of magnet points north, neutral point is obtained on the equatorial line and then

• Referring to the attachment which is according to the Question:-

Therefore,

$B_{E}=B_{H}$

$\large{\frac{\mu}{4 \pi}}{\frac{m}{(r^2 + l^2)^{\frac{3}{2}}}} = B_{H}$

$\large{\frac{10^{-7} \times 0.4}{(0.1^2 + 0.5^2)^{\frac{3}{2}}}}=B_{H}$

$\bf{\therefore B_{H} = 2.86 \times 10^{-5}T}$

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