Points A and B are situated along the extende | vedclass

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

Question

(D) The magnetic field $B$ on the axial line of a short bar magnet is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$,where $d$ is the distance

Vedclass · Accepted Answer

$8 : 1$ approx.

Vedclass · Answer

(D) The magnetic field $B$ on the axial line of a short bar magnet is given by $B = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$,where $d$ is the distance from the center of the magnet.However,the problem specifies the distance $x$ from the nearer pole. Let the length of the magnet be $2l = 2 \ cm$,so $l = 1 \ cm$. The distance from the center is $r = (x + l)$.For a short magnet approximation where $x \gg l$,the magnetic field $B \propto \frac{1}{x^3}$.Thus,the ratio of the magnetic fields at $A$ and $B$ is $\frac{B_A}{B_B} = \left( \frac{2x}{x} \right)^3 = \frac{8}{1}$.Since the magnet length is small compared to the distances,this is an approximate result.

Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be:

Similar Questions

$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

$A$ magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of $60^{\circ}$ and the work done is $W$. The torque on the magnetic needle at this position is

Two identical bar magnets,each of magnetic moment $M$,are kept perpendicular to each other at a certain distance. The magnetic induction at a point that is at the same distance $d$ from the center of both magnets is: (where $\mu_{0}$ is the permeability of free space)

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

The work done in turning a magnet of magnetic moment $M$ by an angle $\theta$ from its initial position parallel to the magnetic meridian is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module