Question

The magnetic field 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 following figures show three cases: In all cases the circular part has radius $r$ and straight ones are infinitely long. For same current, the magnetic field $B$ at the centre $P$ in cases $1,2,3$ has the ratio

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

Answer 1

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

Case:I
Magnetic Fields at $O$:
$B$ due to semi-infinite straight wire $-\left(1\right)$
$B_1=\frac{{\mu }_{0}i}{4\pi r}⨂$

$B$ due to semi-circular wire $-\left(2\right)$
$B_2=\frac{{\mu }_{0}i}{4r}⨀$

$B$ due to semi-infinite wire $-\left(3\right)$
$B_3=\frac{{\mu }_{0}i}{4\pi r}⨀$

Direction of field can be checked by ampere's right hand thumb rule.

So, net magnetic field at $O$ in case $I$ is

$B_I=B_1+B_2+B_3$

$\Rightarrow B_I=-\frac{{\mu }_{0}i}{4\pi r}+\frac{{\mu }_{0}i}{4r}+\frac{{\mu }_{0}i}{4\pi r}=\frac{{\mu }_{0}i}{4r}⨀$

Case:II
Magnetic Fields at $O$:
$B$ due to straight wire- (1) and (3) will be zero as $O$ lies on the axis of these wires

$B$ due to semi-circular wire-(2)
$B_2=\frac{{\mu }_{0}i}{4r}⨂$

Direction of field can be checked by ampere's right hand thumb rule.

Net magnetic field,
$B_{II}=B_1+B_2+B_3=\frac{{\mu }_{0}i}{4r}⨂$

Case:III
Magnetic Fields at $O$:
$B$ due to straight wire (1) will be zero
($0$ lies on its axis)
$B$ due to circular arc (2) will be,

$B_2=\frac{{\mu }_{0}i\theta }{4\pi r}=\frac{{\mu }_{0}i\left(\frac{3\pi }{2}\right)}{4\pi r}⨂$

$r$ is radius of arc,

Angle subtended by arc at centre,
$\theta =\frac{3\pi }{2}$

$B$ due to semi - infinite wire (3) will be
$B_3=\frac{{\mu }_{0}i}{4\pi r}⨀$

Direction of field can be checked by ampere's right hand thumb rule.

Net magnetic field,
$B_{III}=B_1+B_2+B_3=\frac{{\mu }_{0}i}{4\pi r}[\frac{3\pi }{2}-1]⨂$

So, from the expressions of $B_I,B_{II}\&B_{III},$
$\Rightarrow B_1:B_2:B_3=\frac{-\pi }{2}:\frac{\pi }{2}:(\frac{3\pi }{4}-\frac{1}{2})$
(taking inward $+ve$)