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VITEEE 2009 Paper

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Question : 25 of 120

Marks: +1, -0

A bar magnet is 10 cm long is kept with its north (N)-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is

9 A-m
6.75 A-m
27 A-m
1.35 A-m

Solution:

Here, length of magnet = 10 cm = 10 ×

10^{-2}

m, r = 15 ×

10^{-2}

m

OP =

\sqrt{225-25}=\sqrt{200}

cm

Since, at the neutral point, magnetic field due to the magnet is equal to

B_{H}

,

B_{H} = \frac{\mu_{0}}{4\pi} \cdot \frac{M}{(OP^{2}+AO^{2})^{3/2}}

0.4 \times 10^{-4} = 10^{-7} \times \frac{M}{(200 \times 10^{-4} + 25 \times 10^{-4})^{3/2}}

⇒

\frac{0.4 \times 10^{-4}}{10^{-7}} \times (225 \times 10^{-4})^{3/2} = M

⇒

0.4 \times 10^{3} \times 10^{-6} (225)^{3/2} = M

⇒ M = 1.35 A-m

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