A bar magnet of magnetic moment $M$ is divided into ' $n$ ' equal parts by cutting parallel to length. Then one part is suspended in a uniform magnetic field of strength $2 T$ and held making an angle $6 0^{0}$ with the direction of the field. When the magnet is released, the kinetic energy of the magnet in the equilibrium position is:

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Answer 1

Answer is A
Net magnetic moment after cutting $= \frac{M}{n}$
We know that, $∣ U ∣ = M B cos θ$
In equilibrium position, all the potential energy will be converted into kinetic energy i.e. $K E = ∣ U ∣$
Given $M_{n e w} = \frac{M}{n}, B = 2 T, θ = 6 0^{\circ}$
Therefore, $K . E = \frac{M}{n} \times 2 \times cos 6 0^{\circ} = \frac{M}{n} J$

Answer 2

Answer is A
Net magnetic moment after cutting $= \frac{M}{n}$
We know that, $∣ U ∣ = M B cos θ$
In equilibrium position, all the potential energy will be converted into kinetic energy i.e. $K E = ∣ U ∣$
Given $M_{n e w} = \frac{M}{n}, B = 2 T, θ = 6 0^{\circ}$
Therefore, $K . E = \frac{M}{n} \times 2 \times cos 6 0^{\circ} = \frac{M}{n} J$

$\frac{M}{n}$ J

$M n$ J

$\frac{M}{n ^{2}}$ J

$M n^{2}$ J

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A magnet of length $30 c m$ with pole strength $10 A - m$ is freely suspended in a uniform horizontal magnetic field of induction $40 \times 1 0^{- 6} T$ . If the magnet is deflected by $6 0^{o}$ from its equilibrium position, the restoring couple acting on it is :

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Dipole in Uniform Magnetic Field

Solve Question on Magnetic Dipoles

nM​J

Mn J

n2M​J

Mn2 J

Answer is A Net magnetic moment after cutting=nM​ We know that, ∣U∣=MBcosθ In equilibrium position, all the potential energy will be converted into kinetic energy i.e. KE=∣U∣ Given Mnew​=nM​,B=2T,θ=60∘ Therefore, K.E=nM​×2×cos60∘=nM​J

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