(A) Given:Magnetic moment of the bar magnet, $m = 10^{-4} Am^2$Magnetic field, $B = 12 \times 10^{-3} T$Angle, $\theta = 30^{\circ}$The torque $\tau$

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(A) Given:Magnetic moment of the bar magnet, $m = 10^{-4} Am^2$Magnetic field, $B = 12 \times 10^{-3} T$Angle, $\theta = 30^{\circ}$The torque $\tau$ acting on a bar magnet in a uniform magnetic field is given by the formula:$\tau = mB \sin \theta$Substituting the given values:$\tau = (10^{-4} Am^2) \times (12 \times 10^{-3} T) \times \sin(30^{\circ})$Since $\sin(30^{\circ}) = 0.5$:$\tau = 10^{-4} \times 12 \times 10^{-3} \times 0.5$$\tau = 12 \times 10^{-7} \times 0.5$$\tau = 6 \times 10^{-7} Nm$

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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If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

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A
$6 \times 10^{-7} Nm$
B
$6 \times 10^{-5} Nm$
C
$12 \times 10^{-7} Nm$
D
$12 \times 10^{-5} Nm$

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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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If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

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$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \ T$ experiences a torque of magnitude $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment of the magnet will be . . . . . . $J \ T^{-1}$.

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