(A) The torque $\tau$ experienced by a magnetic dipole in a magnetic field is given by $\tau = M B \sin \theta$.For maximum torque,$\theta = 90^\circ$

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(A) The torque $\tau$ experienced by a magnetic dipole in a magnetic field is given by $\tau = M B \sin \theta$.For maximum torque,$\theta = 90^\circ$,so $\tau_{\max} = M B$.Given: $\tau_{\max} = 4 \times 10^{-5} \ N \cdot m$ and $B = 10^{-4} \ Wb/m^2$.Substituting the values: $4 \times 10^{-5} = M \times 10^{-4}$.Solving for $M$: $M = \frac{4 \times 10^{-5}}{10^{-4}} = 4 \times 10^{-1} = 0.4 \ A \cdot m^2$.

Class 12 Physics Magnetism and Matter Bar Magnet and Magnetic Dipole and Magnetic Moment

Easy

$A$ magnet,when placed perpendicular to a uniform magnetic field of strength $10^{-4} \ Wb/m^2$,experiences a maximum torque of $4 \times 10^{-5} \ N \cdot m$. What is its magnetic moment in $A \cdot m^2$?

A
$0.4$
B
$0.2$
C
$0.16$
D
$0.04$

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Magnetism and Matter

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Bar Magnet and Magnetic Dipole and Magnetic Moment

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MediumAIIMS 1995

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$A$ magnet,when placed perpendicular to a uniform magnetic field of strength $10^{-4} \ Wb/m^2$,experiences a maximum torque of $4 \times 10^{-5} \ N \cdot m$. What is its magnetic moment in $A \cdot m^2$?

Similar Questions

$A$ short bar magnet has a length $2l$ and a magnetic moment $10 \text{ Am}^2$. Find the magnetic field at a distance of $z = 0.1 \text{ m}$ from its centre on the axial line. Here,$l$ is negligible as compared to $z$.

$A$ magnet is placed in iron powder and then taken out,then maximum iron powder is at

The force between two identical bar magnets whose centres are $r \ m$ apart is $4.8 \ N$,when their axes are in the same line. If the separation is increased to $2r$,the force between them is reduced to . . . . . . $N$.

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