JEE PYQ: Magnetic Effects Of Current Question 27
Question 27 - 2020 (08 Jan 2020 Shift 2)
A very long wire $ABDMNDC$ is shown in figure carrying current $I$. $AB$ and $BC$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $DMND$ of radius $R$. $AB$, $BC$ parts are tangential to circular turn at $N$ and $D$. Magnetic field at the centre of circle is:
(1) $\frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$
(2) $\frac{\mu_0 I}{2\pi R}\left(\pi - \frac{1}{\sqrt{2}}\right)$
(3) $\frac{\mu_0 I}{2\pi R}(\pi + 1)$
(4) $\frac{\mu_0 I}{2R}$
Show Answer
Answer: (1) $\frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$
Solution
$B_0 = \frac{\mu_0 I}{2R} + \text{contributions from straight wires} = \frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$.
BETA
