(D) The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = m B \sin \theta$. Given: Torque $\t

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$18 \times 10^{-2} \ J \ T^{-1}$

(D) The torque $\tau$ acting on a magnetic dipole in a uniform magnetic field $B$ is given by the formula: $\tau = m B \sin \theta$. Given: Torque $\tau = 4.5 \times 10^{-2} \ J$ Magnetic field $B = 0.5 \ T$ Angle $\theta = 30^{\circ}$ Substituting the values into the formula: $4.5 \times 10^{-2} = m \times 0.5 \times \sin 30^{\circ}$ Since $\sin 30^{\circ} = 0.5$: $4.5 \times 10^{-2} = m \times 0.5 \times 0.5$ $4.5 \times 10^{-2} = m \times 0.25$ $m = \frac{4.5 \times 10^{-2}}{0.25}$ $m = 18 \times 10^{-2} \ J \ T^{-1}$.

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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$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

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A
$3.6 \times 10^2 \ J \ T^{-1}$
B
$36 \times 10^{-2} \ J \ T^{-1}$
C
$1.8 \times 10^2 \ J \ T^{-1}$
D
$18 \times 10^{-2} \ J \ T^{-1}$

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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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$A$ short bar magnet placed with its axis at $30^{\circ}$ to a uniform external magnetic field of $0.5 \ T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \ J$. The magnitude of the magnetic moment is . . . . . . .

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