Question

In the figure $\theta ={tan}^{-1}\left(\sqrt{2}\right)$, then the correct direction of the magnetic field at p can be shown by the arrow

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is A $2$

Given : In the figure $\theta =ta{n}^{-1}\left(\sqrt{2}\right)$ then the correct direction of the magnetic field at P can be show by the arrow

Solution: For M.cosθ point P is at axis, so direction of magnetic field will be along M.cosθ and for M.sinθ point P is perpendicular

So, direction will be just opposite to M Sinθ

Bₐ = $\frac{\mu }{4\pi }\times \frac{2Mcos\theta }{r^3}$

B₁ = $\frac{\mu }{4\pi }\times \frac{2Msin\theta }{r^3}$

tanα = $\frac{B_1}{B_2}$=$\frac{tan\theta }{2}$ ( eq-1)

we have $\theta =ta{n}^{-1}\left(\sqrt{2}\right)$

$tan\theta =\sqrt{2}$

from eq-1 we get $tan\alpha =\frac{\sqrt{2}}{2}$=$\frac{1}{\sqrt{2}}$

$tan\alpha =\frac{1}{tan\theta }$

$tan\alpha =cot\theta$

$tan\alpha =tan(\frac{\pi }{2}-\theta )$

$\alpha =(\frac{\pi }{2}-\theta )$

$\theta +\theta =\frac{\pi }{2}$

Thus the correct direction of the magnetic field is $\theta =ta{n}^{-1}\left(\sqrt{2}\right)$

The correct opt= A