The ratio of magnetic field at the centre of a current carrying coil of radius $r$ to the magnetic field at distance $r$ from the centre of coil on its axis is $\sqrt{x}: 1$. The value of $x$ is __________
Answer (integer)
8
Solution
<p>The magnetic field at the center of a loop (B<sub>1</sub>) is given by</p>
<p>$ B_1 = \frac{\mu_0 I}{2r} $</p>
<p>The magnetic field on the axis of the loop at a distance ( r ) from the center (B<sub>2</sub>) is given by</p>
<p>$ B_2 = \frac{\mu_0 Ir^2}{2(r^2 + d^2)^{3/2}} $</p>
<p>where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get</p>
<p>$ B_2 = \frac{\mu_0 I}{4\sqrt{2}r} $</p>
<p>The ratio of $ B_1 $ to $ B_2 $ is</p>
<p>$ \frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \sqrt{8} : 1 $</p>
<p>So, the value of ( x ) is 8.</p>
About this question
Subject: Physics · Chapter: Magnetic Effects of Current · Topic: Biot-Savart Law
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