Question

A wire shown in figure carries a current of $40A$ . If $r=3.14cm$ , the magnetic field at point $P$ will be :

• A. $1.6\times 10^{-3}T$
• B. $3.2\times 10^{-2}T$
• C. $4.8\times 10^{-3}T$
• D. $6.0\times 10^{-4}T$

Answer 1

Answer: (D)

Let the given circular $ABC$ part of wire subtends an angle $\theta$ at its centre. Then, magnetic field due to this circular part is

$B^{\prime }=B_c\times \frac{\theta }{2\pi }=\frac{{\mu }_0}{4\pi }\times \frac{2\pi i}{e}\times \frac{\theta }{2\pi }$
$\Rightarrow B^{\prime }=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}\theta$
Given, $i=40A,r=3.14cm=3.14\times 10^{-2}m$
$\theta =360^{\circ }-90^{\circ }=270^{\circ }=\frac{3\pi }{2}rad$
$\therefore B^{\prime }=\frac{10^{-7}\times 40}{3.14\times 10^{-2}}\times \frac{3\pi }{2}=60\times 10^{-5}$
$B^{\prime }=6\times 10^{-4}T$