Two magnets A and B are identical and are arr | vedclass

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

Question

(C) The magnetic field due to magnet $A$ (axial position) at point $P$ is $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_1^3}$.The magnetic field due to

Vedclass · Accepted Answer

$(2\cot 	heta )^{1/3}$

Vedclass · Answer

(C) The magnetic field due to magnet $A$ (axial position) at point $P$ is $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_1^3}$.The magnetic field due to magnet $B$ (equatorial position) at point $P$ is $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{d_2^3}$.When the magnetic needle is in equilibrium at an angle $	heta$ with the direction of $B_1$,the resultant magnetic field makes an angle $	heta$ with $B_1$,such that $	an 	heta = \frac{B_2}{B_1}$.Substituting the values,we get $	an 	heta = \frac{(\mu_0 / 4\pi) \cdot (M / d_2^3)}{(\mu_0 / 4\pi) \cdot (2M / d_1^3)} = \frac{d_1^3}{2d_2^3}$.Therefore,$\frac{d_1^3}{d_2^3} = 2 	an 	heta$,which implies $\frac{d_1}{d_2} = (2 	an 	heta)^{1/3}$.Wait,checking the equilibrium condition: The needle aligns with the resultant field. If $	heta$ is the deflection from the axis of magnet $A$ (field $B_1$),then $	an 	heta = B_2 / B_1$. Thus $\frac{d_1^3}{2d_2^3} = 	an 	heta \Rightarrow \frac{d_1}{d_2} = (2 	an 	heta)^{1/3}$.Re-evaluating the provided options and standard convention: If $	heta$ is the angle with the vertical (field $B_2$),then $	an 	heta = B_1 / B_2 = \frac{2d_2^3}{d_1^3} \Rightarrow \frac{d_1^3}{d_2^3} = 2 \cot 	heta \Rightarrow \frac{d_1}{d_2} = (2 \cot 	heta)^{1/3}$. Based on the figure,$	heta$ is the angle with the vertical,so option $C$ is correct.

Similar Questions

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the quarter circle of radius $a$ and center at the origin in the first quadrant of the $x-z$ plane,closed by segments along the $x$ and $z$ axes.

$A$ bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it,then the angle by which it is to be rotated is....$^o$

$A$ short bar magnet of magnetic moment $2.5 \text{ Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \text{ T}$. The work done in moving the magnet from its most stable position to its most unstable position is:

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module