(A) Given, magnetic moment $M = 200 \text{ A m}^2$. Magnetic field $B = 0.30 \text{ N A}^{-1} \text{ m}^{-1}$. Angle $\theta = 30^{\circ}$. We know th

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(A) Given, magnetic moment $M = 200 \text{ A m}^2$. Magnetic field $B = 0.30 \text{ N A}^{-1} \text{ m}^{-1}$. Angle $\theta = 30^{\circ}$. We know that the torque $\tau$ acting on a magnetic dipole in a magnetic field is given by the formula: $\tau = M B \sin \theta$. Substituting the given values: $\tau = 200 \times 0.30 \times \sin(30^{\circ})$. Since $\sin(30^{\circ}) = \frac{1}{2}$,$\tau = 200 \times 0.30 \times \frac{1}{2} = 100 \times 0.30 = 30 \text{ N m}$. Thus, the torque required is $30 \text{ N m}$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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

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A
$30 \text{ N m}$
B
$30 \sqrt{3} \text{ N m}$
C
$60 \text{ N m}$
D
$60 \sqrt{3} \text{ N m}$

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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

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$A$ short bar magnet is placed in a uniform magnetic field of $2 \ T$ such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36 \sqrt{2} \ Nm$,then the magnetic moment of the magnet is: (in $J \ T^{-1}$)

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