$\mathrm{B}_{X}$ and $\mathrm{B}_{\mathrm{Y}}$ are the magnetic fields at the centre of two coils $\mathrm{X}$ and $\mathrm{Y}$ respectively each carrying equal current. If coil $X$ has 200 turns and $20 \mathrm{~cm}$ radius and coil $Y$ has 400 turns and $20 \mathrm{~cm}$ radius, the ratio of $B_{X}$ and $B_{Y}$ is :
Solution
<p>$B = {{{\mu _0}NI} \over {2R}}$</p>
<p>${{{B_X}} \over {{B_Y}}} = {{{N_x}{R_y}} \over {{N_y}{R_x}}}$</p>
<p>$= {{200 \times 20} \over {400 \times 20}} = {1 \over 2}$</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Magnetic Effects of Current are available — start with the harder ones if your accuracy is >70%.