Two short magnets are placed along the same a | vedclass

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

Question

(D) Both magnets are placed in the magnetic field of one another. The magnetic field $B_1$ produced by magnet $1$ at the position of magnet $2$ (on it

Vedclass · Accepted Answer

Fourth power of the distance

Vedclass · Answer

(D) Both magnets are placed in the magnetic field of one another. The magnetic field $B_1$ produced by magnet $1$ at the position of magnet $2$ (on its axial line) is given by:$B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M_1}{r^3}$The potential energy $U$ of a magnetic dipole $M_2$ in an external magnetic field $B_1$ is given by $U = -M_2 B_1 \cos 	heta$. Since the magnets are placed along the same axis with like poles facing each other,the magnetic moments are aligned,so $	heta = 0^\circ$ and $\cos 0^\circ = 1$. Thus:$U = -M_2 B_1 = -M_2 \left( \frac{\mu_0}{4\pi} \cdot \frac{2M_1}{r^3} ight) = -\frac{\mu_0}{4\pi} \cdot \frac{2M_1 M_2}{r^3}$The force $F$ between the magnets is given by the negative gradient of the potential energy:$F = -\frac{dU}{dr} = -\frac{d}{dr} \left( -\frac{\mu_0}{4\pi} \cdot \frac{2M_1 M_2}{r^3} ight)$$F = \frac{\mu_0}{4\pi} \cdot 2M_1 M_2 \cdot \frac{d}{dr} (r^{-3}) = \frac{\mu_0}{4\pi} \cdot 2M_1 M_2 \cdot (-3r^{-4})$$F = -\frac{\mu_0}{4\pi} \cdot \frac{6M_1 M_2}{r^4}$The magnitude of the force is $F = \frac{\mu_0}{4\pi} \cdot \frac{6M_1 M_2}{r^4}$.Therefore,the force varies inversely as the fourth power of the distance,i.e.,$F \propto \frac{1}{r^4}$.

Two short magnets are placed along the same axis with their like poles facing each other. They repel each other with a force which varies inversely as:

Similar Questions

$A$ 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 $...... \times 10^{-4} \, N \cdot m$.

$A$ short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of $0.16 \ T$ experiences a torque of magnitude $0.032 \ J$. The magnetic moment of the bar magnet will be......$J/T$.

Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module