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:

Question

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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

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 :

If a bar magnet in magnetic moment $m$ is deflected from an angle $θ$ in a uniform magnetic field of induction $B$ , the work done in reversing the direction is

A bar magnet has a magnetic moment of $200$ $A m^{2}$ . The magnet is suspended in a magnetic field of $0.30 N A^{- 1} m^{- 1}$ . The torque required to rotate the magnet from its equilibrium position through an angle of $30^{o}$ will be:

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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

A short bar magnet of magnetic moment m=0.32JT−1 is placed in a uniform magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

A magnet of length 30 cm with pole strength 10 A−m is freely suspended in a uniform horizontal magnetic field of induction 40×10−6T . If the magnet is deflected by 60o from its equilibrium position, the restoring couple acting on it is :

If a bar magnet in magnetic moment m is deflected from an angle θ in a uniform magnetic field of induction B, the work done in reversing the direction is

A bar magnet has a magnetic moment of 200 Am2. The magnet is suspended in a magnetic field of 0.30 NA−1m−1. The torque required to rotate the magnet from its equilibrium position through an angle of 30o will be: