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Byju's Answer

Standard XII

Physics

Magnetic Flux and Faraday's Law

Two short bar...

Question

Two short bar magnets of magnetic moments $0.125 A m^{2}$ and $0.512 A m^{2}$ are placed with their like poles facing each other. If the distance between the centres of the magnet is $0.26 m$ . The distance of neutral point from the weaker magnet is

0.13 m

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0.2 m

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0.26 m

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0.1 m

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Solution

The correct option is A $0.1 m$
$B_{a x i a l} = \frac{μ_{0} 2 M}{4 π r^{3}}$
Equating the magnetic fields due to the two bar magnets, $\frac{M_{1}}{x^{3}} = \frac{M_{2}}{{(d - x)}^{3}}$
Solving the above for x (distance from the center of weaker magnet to the neutral point), gives the formula in the hint. Putting in the values, we get, $x = 0.1 m$

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Disclaimer: No option as

0.08 d y n e

was given in the original paper.

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Two short bar magnets of magnetic moments 0.125Am2 and 0.512Am2 are placed with their like poles facing each other. If the distance between the centres of the magnet is 0.26m. The distance of neutral point from the weaker magnet is

The correct option is A 0.1mBaxial=μ02M4πr3Equating the magnetic fields due to the two bar magnets, M1x3=M2(d−x)3Solving the above for x (distance from the center of weaker magnet to the neutral point), gives the formula in the hint. Putting in the values, we get, x=0.1m

Two short bar magnets of magnetic moments $0.125 A m^{2}$ and $0.512 A m^{2}$ are placed with their like poles facing each other. If the distance between the centres of the magnet is $0.26 m$ . The distance of neutral point from the weaker magnet is