Question

A circular ring is connected across a cell such that the current $I$, enters the loop at point $A$ and leaves at point $B$, as shown in the diagram. The radius of the ring is $r$. The magnetic field at the centre of the ring is

• A. Zero
• B. $\frac{{\mu }_{o}I}{4r}$
• C. $\frac{{\mu }_{o}I}{2r}$
• D. $\frac{{\mu }_{o}I}{r}$

Answer 1

The correct option is A Zero

Given, the smaller arc subtends an angle $\theta$ at the centre of the ring, so that the bigger arc subtends an angle $(\pi -\theta$) at the centre.

Let, the resistance of the smaller arc is $R_1$ and that of bigger arc is $R_2$.

Now, $V_{AB}=I_1{R}_1=I_2{R}_2....\left(i\right)$

Here, $R_1=\frac{\rho l_1}{A},R_2=\frac{\rho l_2}{A}$

Now, $l_1=r\theta$ and $l_2=r(\pi -\theta )$

$\therefore R_1=\frac{\rho r\theta }{A}....\left(ii\right)$ and

$R_2=\frac{\rho r(\pi -\theta )}{A}....\left(iii\right)$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we can write,

$I_1\frac{\rho r\theta }{A}=I_2\frac{\rho r(\pi -\theta )}{A}$

$\therefore I_1\theta =I_2(\pi -\theta )....\left(iv\right)$

Magnetic fields at the centre due to arcs are,

$B_1=\frac{{\mu }_o{I}_1}{4\pi r^2}{l}_1=\frac{{\mu }_o{I}_1}{4\pi r^2}r\theta$

$B_2=\frac{{\mu }_o{I}_2}{4\pi r^2}{l}_2=\frac{{\mu }_o{I}_1}{4\pi r^2}r(\pi -\theta )$

As, $I_1\theta =I_2(\pi -\theta )$

$\Rightarrow B_1=B_2$

Since the directions of $I_1$ and $I_2$ are opposite, so from the "Fleming's right-hand thumb rule", the directions of ${\overrightarrow{B}}_1$ and ${\overrightarrow{B}}_2$ will also be opposite.

So, the net magnetic field at the centre will be,

$B_{net}=B_1-B_2=0$

Hence, option $\left(a\right)$ is the correct answer.