A small magnetized needle P is placed at poin | vedclass

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

Question

(C) magnetic dipole is in equilibrium in an external magnetic field when its magnetic moment $\vec{m}$ is parallel or anti-parallel to the magnetic fi

Vedclass · Accepted Answer

$PQ_1, PQ_2$

Vedclass · Answer

(C) magnetic dipole is in equilibrium in an external magnetic field when its magnetic moment $\vec{m}$ is parallel or anti-parallel to the magnetic field $\vec{B}$ (i.e.,$	heta = 0^{\circ}$ or $180^{\circ}$).At point $O$,the magnetic needle $P$ has a magnetic moment pointing upwards.For any magnet $Q$ placed at a position,it experiences a magnetic field $\vec{B}$ due to the dipole $P$. The system is in equilibrium if the magnetic moment $\vec{m}_Q$ of magnet $Q$ is parallel or anti-parallel to the magnetic field $\vec{B}$ produced by $P$ at that location.$1$. At $Q_1$ and $Q_2$ (axial positions),the magnetic field $\vec{B}$ is parallel to the magnetic moment of $P$. However,the orientation of $Q_1$ and $Q_2$ is horizontal,while the field is vertical. Thus,they are not in equilibrium.$2$. At $Q_3, Q_4, Q_5, Q_6$ (equatorial positions),the magnetic field $\vec{B}$ is anti-parallel to the magnetic moment of $P$. The orientation of these magnets is vertical,which is parallel/anti-parallel to the field,so they are in equilibrium.Therefore,the system is not in equilibrium for configurations $PQ_1$ and $PQ_2$.

$A$ 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?

Similar Questions

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

The magnitude of the axial field due to a short bar magnet at a distance of $50 \,cm$ from its mid-point is (The magnetic moment of the bar magnet is $0.4 \,Am^2$)

$A$ magnet of length $0.1 \, m$ and pole strength $10^{-4} \, A \cdot m$ is kept in a magnetic field of $30 \, Wb/m^2$ at an angle of $30^\circ$. The couple (torque) acting on it is $...... \times 10^{-4} \, N \cdot m$.

$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is:

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module