Question

Two short magnets each of moment $M$ are placed one over the other at right angles. The resultant magnetic induction at a point $\text{P}$ that lies at a distance $d$ from the common centre is-

• A. $\left(\frac{{\mu }_0}{4\pi }\right)\frac{M}{d^3}$
• B. $\left(\frac{{\mu }_0}{4\pi }\right)\frac{3M}{d^3}$
• C. $\sqrt{3}\times \left(\frac{{\mu }_0}{4\pi }\right)\frac{M}{d^3}$
• D. $\sqrt{5}\times \left(\frac{{\mu }_0}{4\pi }\right)\frac{M}{d^3}$
  
  <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->

Answer 1

The correct option is D $\sqrt{5}\times \left(\frac{{\mu }_0}{4\pi }\right)\frac{M}{d^3}$


<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
As we can see, $\text{P}$ is an equatorial and axial point with respect to magnet $\left(1\right)$ and $\left(2\right)$ respectively.



The resultant field at point $\text{P}$ is,

$B_R=\sqrt{(B_{eq}{)}_1^2+(B_{ax}{)}_2^2}$

$=\sqrt{(B_{eq}{)}_1^2+(2B_{eq}{)}_1^2}[\because B_{ax}=2B_{eq}]$

$=\sqrt{5}{\left(B_{eq}\right)}_1$

$=\sqrt{5}\times \left(\frac{{\mu }_0}{4\pi }\right)\frac{M}{d^3}$

Hence, $\left(D\right)$ is the correct answer.