A bar magnet has a magnetic moment $2.5 J T^{- 1}$ and is placed in a magnetic field of $0.2 T$ . Work done in turning the magnet from parallel to anti-parallel position relative to the field direction.

0.5 J

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1 J

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2.0 J

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Zero

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Solution

The correct option is B $1 J$
$P E = M B (1 - cos θ)$
so $P E_{1} = M B (1 - cos θ) = 0$
$P E_{2} = M B (1 - cos 180^{o}) = 2 M B = 2 \times \frac{2.5 J}{7} \times 0.27$
$P . E_{2} = 1$ Joule
$w e r e = P . E_{2} - P . E_{1} = 1.0 = 1 J o u l e$

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A bar magnet has a magnetic moment 2.5JT−1 and is placed in a magnetic field of 0.2 T. Work done in turning the magnet from parallel to anti-parallel position relative to the field direction.

The correct option is B 1 JPE=MB(1−cosθ)so PE1=MB(1−cosθ)=0PE2=MB(1−cos180o)=2MB=2×2.5 J7×0.27P.E2=1 Joulewere=P.E2−P.E1=1.0=1 Joule

A bar magnet has a magnetic moment $2.5 J T^{- 1}$ and is placed in a magnetic field of $0.2 T$ . Work done in turning the magnet from parallel to anti-parallel position relative to the field direction.

The correct option is B $1 J$
$P E = M B (1 - cos θ)$
so $P E_{1} = M B (1 - cos θ) = 0$
$P E_{2} = M B (1 - cos 180^{o}) = 2 M B = 2 \times \frac{2.5 J}{7} \times 0.27$
$P . E_{2} = 1$ Joule
$w e r e = P . E_{2} - P . E_{1} = 1.0 = 1 J o u l e$