A bar magnet,when placed at an angle of 30^o | vedclass

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

Question

(A) The torque $	au$ experienced by a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole moment

Vedclass · Accepted Answer

$2 	imes 10^{-2} \,A-m$

Vedclass · Answer

(A) The torque $	au$ experienced by a magnetic dipole in a magnetic field is given by $	au = MB \sin 	heta$,where $M$ is the magnetic dipole moment,$B$ is the magnetic field,and $	heta$ is the angle.Since $M = m 	imes L$ (where $m$ is the pole strength and $L$ is the length of the magnet),the formula becomes $	au = (m 	imes L) B \sin 	heta$.Given: $	au = 25 	imes 10^{-6} \,N-m$,$B = 5 	imes 10^{-2} \,T$,$L = 5 \,cm = 5 	imes 10^{-2} \,m$,and $	heta = 30^o$.Substituting the values: $25 	imes 10^{-6} = (m 	imes 5 	imes 10^{-2}) 	imes (5 	imes 10^{-2}) 	imes \sin 30^o$.Since $\sin 30^o = 0.5$,we have $25 	imes 10^{-6} = m 	imes 25 	imes 10^{-4} 	imes 0.5$.$25 	imes 10^{-6} = m 	imes 12.5 	imes 10^{-4}$.$m = \frac{25 	imes 10^{-6}}{12.5 	imes 10^{-4}} = 2 	imes 10^{-2} \,A-m$.

$A$ bar magnet,when placed at an angle of $30^o$ to the direction of a magnetic field of induction $5 \times 10^{-2} \,T$,experiences a torque of $25 \times 10^{-6} \,N-m$. If the length of the magnet is $5 \,cm$,what is its pole strength?

Similar Questions

The work done in rotating a bar magnet,which is initially in the direction of a uniform magnetic field,through $45^{\circ}$ is $W$. The additional work to be done to rotate the magnet further through $15^{\circ}$ is

$A$ magnet of magnetic moment $M$ is situated with its axis along the direction of a magnetic field of strength $B$. The work done in rotating it by an angle of $180^{\circ}$ will be

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.

The intensity of the magnetic field is $H$ and the magnetic moment of the magnet is $M$. The maximum potential energy is $..... MH$.

$A$ 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module