Two identical short bar magnets,each having m | vedclass

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

Question

(D) Let the two magnets be $1$ and $2$. The point $P$ is at a distance $d$ from the center of each magnet.For magnet $1$,point $P$ lies on its axial l

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} (\sqrt{5}) \frac{M}{d^3}$

Vedclass · Answer

(D) Let the two magnets be $1$ and $2$. The point $P$ is at a distance $d$ from the center of each magnet.For magnet $1$,point $P$ lies on its axial line. The magnetic field $B_1$ due to magnet $1$ at $P$ is given by $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3}$.For magnet $2$,point $P$ lies on its equatorial line. The magnetic field $B_2$ due to magnet $2$ at $P$ is given by $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3}$.Since the axes are perpendicular,the fields $B_1$ and $B_2$ are perpendicular to each other.The net magnetic field $B_{net}$ at point $P$ is $B_{net} = \sqrt{B_1^2 + B_2^2}$.Substituting the values: $B_{net} = \sqrt{\left( \frac{\mu_0}{4\pi} \cdot \frac{2M}{d^3} \right)^2 + \left( \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \right)^2}$.$B_{net} = \frac{\mu_0}{4\pi} \cdot \frac{M}{d^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4\pi} \cdot \frac{\sqrt{5}M}{d^3}$.

Two identical short bar magnets,each having magnetic moment $M$,are placed a distance of $2d$ apart with their axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is:

Similar Questions

How do ortho and para hydrogen differ from each other?

If $3 \hat{i}+3 \hat{j}+\sqrt{3} \hat{k}$, $\hat{i}+\hat{k}$, and $\sqrt{3} \hat{i}+\sqrt{3} \hat{j}+\lambda \hat{k}$ are coplanar, then $\lambda$ is equal to

Which of the following are not electromagnetic waves?

The following compound is used as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module