Two identical short bar magnets,each having a | vedclass

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

Question

(D) Let the two magnets be $M_1$ and $M_2$ with magnetic moment $M = 10 \, Am^2$. The distance between their centres is $0.2 \, m$,so the distance fro

Vedclass · Accepted Answer

$\sqrt{5} 	imes 10^{-3} \, T$

Vedclass · Answer

(D) Let the two magnets be $M_1$ and $M_2$ with magnetic moment $M = 10 \, Am^2$. The distance between their centres is $0.2 \, m$,so the distance from the centre of each magnet to the midpoint $P$ is $d = 0.1 \, m$.For magnet $1$,point $P$ lies on its axial line. The magnetic field $B_a$ is given by $B_a = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$.For magnet $2$,point $P$ lies on its equatorial line. The magnetic field $B_e$ is given by $B_e = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3}$.Since these fields are perpendicular,the resultant magnetic field is $B_{net} = \sqrt{B_a^2 + B_e^2}$.$B_{net} = \sqrt{\left( \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} ight)^2 + \left( \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} ight)^2} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{5}$.Substituting the values: $B_{net} = 10^{-7} \cdot \frac{10}{(0.1)^3} \cdot \sqrt{5} = 10^{-7} \cdot \frac{10}{0.001} \cdot \sqrt{5} = 10^{-7} \cdot 10^4 \cdot \sqrt{5} = \sqrt{5} 	imes 10^{-3} \, T$.

Similar Questions

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.25 \; T$ experiences a torque of magnitude equal to $4.5 \times 10^{-2} \; J$. What is the magnitude of magnetic moment (in $J \; T^{-1}$) of the magnet?

Some equipotential surfaces of the magnetic scalar potential are shown in the figure. The magnetic field at a point in the region is

$A$ magnet of magnetic moment $M$ is lying in a uniform magnetic field $B$. $W_1$ is the work done in turning it from $0^o$ to $60^o$ and $W_2$ is the work done in turning it from $30^o$ to $90^o$. Then:

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module