Question

A vertical circular coil of one turn and radius 9.42 cm is placed with its plane in the magnetic meridian and short magnetic needle is pivoted at the centre of the coil so that it can freely rotate in the horizontal plane. If a current of 6A is passed through the coil, then the needle deflects by $(B_H=4\times {10}^{-5}T)$

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

The needle at the centre gets deflected and comes to an equilibrium position under the action of two perpendicular field$:$

one due to horizontal component of earth and the other due to field$\left(B\right)$ set up the coil due to current$.$

$B=B_H\tan \theta$

and $B=\frac{{\mu }_{0}ni}{2r}$

$n=no.ofterms=1,i=6A,r=9.42cmand{B}_{{}^H}=4\times {10}^{-5}T$

therefore$,$

$B=B_H\tan \theta =\frac{{\mu }_{0}ni}{2r}$

$\tan \theta =\frac{{\mu }_{0}ni}{2r{B}_H}=\frac{4\pi \times {10}^{-7}\times 1\times 6}{2\times 9.42\times {10}^{-2}\times 4\times {10}^{-5}}=1$

$\theta ={45}^0$

hence,

option $\left(B\right)$ is correct answer.