A short magnetic needle is placed in a magnet | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The torque $\vec{	au}$ on a magnetic dipole $\vec{M}$ in a magnetic field $\vec{B}$ is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.Let the mag

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(A) The torque $\vec{	au}$ on a magnetic dipole $\vec{M}$ in a magnetic field $\vec{B}$ is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.Let the magnitude of the magnetic moment be $M$. The unit vector in the direction of the needle is $\hat{u} = \frac{\sqrt{3}\hat{i} + \hat{j}}{2}$. Thus,$\vec{M} = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j})$.Given $\vec{B}_1 = B\hat{i}$,the torque is $\vec{	au}_1 = \frac{M}{2}(\sqrt{3}\hat{i} + \hat{j}) 	imes B\hat{i} = \frac{MB}{2}(\sqrt{3}(\hat{i} 	imes \hat{i}) + (\hat{j} 	imes \hat{i})) = \frac{MB}{2}(0 - \hat{k}) = -\frac{MB}{2}\hat{k}$.Given $|\vec{	au}_1| = 0.06 \ N-m$,we have $\frac{MB}{2} = 0.06$,so $MB = 0.12 \ N-m$.In the second case,the needle is oriented in the direction $(\hat{i} + \sqrt{3}\hat{j})$,so $\vec{M}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j})$. The magnetic field is $\vec{B}_2 = 2B\hat{j}$.The torque is $\vec{	au}_2 = \vec{M}_2 	imes \vec{B}_2 = \frac{M}{2}(\hat{i} + \sqrt{3}\hat{j}) 	imes 2B\hat{j} = MB(\hat{i} 	imes \hat{j} + \sqrt{3}(\hat{j} 	imes \hat{j})) = MB(\hat{k} + 0) = MB\hat{k}$.Thus,$|\vec{	au}_2| = MB = 0.12 \ N-m$.

Similar Questions

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$, find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

The work done in rotating a magnet of magnetic moment $2 \, A-m^2$ in a magnetic field of $5 \times 10^{-3} \, T$ from the direction along the magnetic field to the opposite direction to the magnetic field is:

$A$ short bar magnet of magnetic moment $0.4\,JT^{-1}$ is placed in a uniform magnetic field of $0.16\,T$. The magnet is in stable equilibrium when the potential energy is

$A$ 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 $J \ T^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module