Two identical dipoles each of magnetic moment | vedclass

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

Question

(A) Given: Magnetic moment $M = 1.0 \, A-m^2$,distance between dipoles $d = 2 \, m$. The point is midway,so the distance from each dipole to the point

Vedclass · Accepted Answer

$\sqrt{5} 	imes 10^{-7} \, T$

Vedclass · Answer

(A) Given: Magnetic moment $M = 1.0 \, A-m^2$,distance between dipoles $d = 2 \, m$. The point is midway,so the distance from each dipole to the point is $r = 1 \, m$.For the first dipole,the point lies on its axial line. The magnetic field is $B_1 = \frac{\mu_0}{4\pi} \frac{2M}{r^3} = 10^{-7} 	imes \frac{2 	imes 1}{1^3} = 2 	imes 10^{-7} \, T$.For the second dipole,the point lies on its equatorial line. The magnetic field is $B_2 = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 10^{-7} 	imes \frac{1}{1^3} = 1 	imes 10^{-7} \, T$.Since the axes are perpendicular,the fields $B_1$ and $B_2$ are perpendicular to each other.The resultant magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{(2 	imes 10^{-7})^2 + (1 	imes 10^{-7})^2} = \sqrt{4 + 1} 	imes 10^{-7} = \sqrt{5} 	imes 10^{-7} \, T$.

Two identical dipoles each of magnetic moment $1.0 \, A-m^2$ are placed at a separation of $2 \, m$ with their axes perpendicular to each other. What is the magnetic field at a point midway between the dipoles?

Similar Questions

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:

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is:

$A$ 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?

$A$ magnet of magnetic moment $2 \, J \, T^{-1}$ is aligned in the direction of a magnetic field of $0.1 \, T$. What is the net work done to bring the magnet normal to the magnetic field?

Vedclass Test Series

Exam Paper Generator

Online Exam Module