Question

The field B at the centre of a circular coil of radius $r$ is $\pi$ times that due to a long straight wire at a distance $r$ from it for equal currents. The figure shows three cases. In all cases the circular part has radius $r$ and straight ones are infinitely long. For same current, the field B at the centre P in cases 1, 2, 3 has the ratio:

• A. $(-\frac{\pi }{2}):\frac{\pi }{2}:(\frac{3\pi }{4}-\frac{1}{2})$
• B. $(-\frac{\pi }{2}+1):(\frac{\pi }{2}+1):(\frac{3\pi }{4}-\frac{1}{2})$
• C. $-\frac{\pi }{2}:\frac{\pi }{2}:\frac{3\pi }{4}$
• D. $(-\frac{\pi }{2}-1):(\frac{\pi }{4}+\frac{1}{4}):(\frac{3\pi }{4}+\frac{1}{2})$

Answer 1

The correct option is A $(-\frac{\pi }{2}):\frac{\pi }{2}:(\frac{3\pi }{4}-\frac{1}{2})$

For case (a) magnetic field due to straight portions is cancelled and the magnetic field due to semi circular arc of radius r at P is:

$B_a=\frac{1}{2}\times \frac{{\mu }_{0}i}{2r}=\left(\frac{{\mu }_{0}i}{4\pi r}\right)\times \pi$
It is in upward direction and we take upward direction negative, So:
${\overrightarrow{B}}_a=-\left(\frac{{\mu }_{0}i}{4\pi r}\right)\pi$

For case (b) Due to straight portion, the magnetic field is zero. So the magnetic field due to semi circular arc is:

${\overrightarrow{B}}_b=\left(\frac{{\mu }_{0}i}{4\pi r}\right)\times \pi$ (in downwards direction so + iv sign)

For case (c) Magnetic field due to straight portion is $=-\frac{{\mu }_0}{4\pi }\frac{i}{r}$ (upward direction)

Magnetic field due to circular arc which make an angle $3\pi /2$ at centre is$=\left(\frac{{\mu }_{0}i}{4\pi r}\right)\times \frac{3\pi }{2}$ (downward direction)

So ${\overrightarrow{B}}_c=\left(\frac{{\mu }_{0}i}{4\pi r}\right)(\frac{3\pi }{2}-1)$

Then the ratio: ${\overrightarrow{B}}_a:{\overrightarrow{B}}_b:{\overrightarrow{B}}_c=-\pi :\pi :(\frac{3\pi }{2}-1)$$=\left(\frac{-\pi }{2}\right):\frac{\pi }{2}:(\frac{3\pi }{4}-\frac{1}{2})$