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Standard XII

Physics

Magnetic Moment

A bar magnet ...

Question

A bar magnet of magnetic moment $3.0 A - m^{2}$ is placed in a uniform magnetic induction field of $2 \times 10^{-} 5 T$ . If each pole of the magnetic experiences a force of $6 \times 10^{-} 4 N$ , the length of the magnet is:

0.5 m

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0.3 m

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0.2 m

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0.1 m

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Solution

The correct option is D $0.1 m$
Magnetic moment is

$M = m L$

$m = \frac{M}{L}$

$∵ F = m B$

$F = \frac{M}{L} B$

$L = \frac{M B}{F}$

$L = \frac{3 \times 2 \times 10^{- 5}}{6 \times 10^{- 4}}$

$L = 0.1 m$
Hence, Length of magnet is $0.1 m$

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A bar magnet of magnetic moment 3.0A−m2 is placed in a uniform magnetic induction field of 2×10−5T. If each pole of the magnetic experiences a force of 6×10−4N, the length of the magnet is:

The correct option is D 0.1 mMagnetic moment is M=mL m=ML ∵F=mB F=MLB L=MBF L=3×2×10−56×10−4 L=0.1m Hence, Length of magnet is 0.1m

A bar magnet of magnetic moment $3.0 A - m^{2}$ is placed in a uniform magnetic induction field of $2 \times 10^{-} 5 T$ . If each pole of the magnetic experiences a force of $6 \times 10^{-} 4 N$ , the length of the magnet is:

The correct option is D $0.1 m$
Magnetic moment is

$M = m L$

$m = \frac{M}{L}$

$∵ F = m B$

$F = \frac{M}{L} B$

$L = \frac{M B}{F}$

$L = \frac{3 \times 2 \times 10^{- 5}}{6 \times 10^{- 4}}$

$L = 0.1 m$
Hence, Length of magnet is $0.1 m$