Two short bar magnets, each with a magnetic m | vedclass

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

Question

(C) The origin is on the axial line of the first magnet $(M_1)$ and on the equatorial line of the second magnet $(M_2)$.Given: $M = 9 	ext{ Am}^2$, $

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(C) The origin is on the axial line of the first magnet $(M_1)$ and on the equatorial line of the second magnet $(M_2)$.Given: $M = 9 	ext{ Am}^2$, $r = 3 	ext{ cm} = 3 	imes 10^{-2} 	ext{ m}$.Magnetic field due to $M_1$ (axial point) at the origin:$B_1 = \frac{\mu_0}{4 \pi} 	imes \frac{2M}{r^3} = 10^{-7} 	imes \frac{2 	imes 9}{(3 	imes 10^{-2})^3} = 10^{-7} 	imes \frac{18}{27 	imes 10^{-6}} = \frac{2}{3} 	imes 10^{-1} 	ext{ T}$.This field is directed along the positive $X$-axis.Magnetic field due to $M_2$ (equatorial point) at the origin:$B_2 = \frac{\mu_0}{4 \pi} 	imes \frac{M}{r^3} = 10^{-7} 	imes \frac{9}{(3 	imes 10^{-2})^3} = 10^{-7} 	imes \frac{9}{27 	imes 10^{-6}} = \frac{1}{3} 	imes 10^{-1} 	ext{ T}$.Since the magnetic moment of $M_2$ is in the negative $X$-direction, the equatorial field at the origin points in the positive $X$-direction.As both $B_1$ and $B_2$ point in the same direction, the resultant magnetic field is:$B = B_1 + B_2 = \left(\frac{2}{3} + \frac{1}{3}ight) 	imes 10^{-1} 	ext{ T} = 1 	imes 10^{-1} 	ext{ T} = 0.1 	ext{ T}$.

Similar Questions

If a bar magnet of moment $10^{-4} Am^2$ is kept in a uniform magnetic field of $12 \times 10^{-3} T$ such that it makes an angle of $30^{\circ}$ with the direction of the magnetic field, then the torque acting on the magnet is:

Give the stability position of a bar magnet for $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$.

$A$ bar magnet has a magnetic moment of $200 \text{ A m}^2$. The magnet is suspended in a magnetic field of $0.30 \text{ N A}^{-1} \text{ m}^{-1}$. The torque required to rotate the magnet from its equilibrium position through an angle of $30^{\circ}$ will be:

Two short magnets of equal dipole moments $M$ are fastened perpendicularly at their centre (figure). The magnitude of the magnetic field at a distance $d$ from the centre on the bisector of the right angle is

$A$ 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module