Question

A battery is connected between two points $\text{A}$ and $\text{B}$. On the circumference of a conducting ring of radius $r$ and resistance $R$. One of the arcs $\text{ACB}$ of the ring substends an angle $\theta$ at the centre. The value of magnetic induction at the centre due to current in the ring is

• A. Proportional to $2(180-\theta )$
  
• B. Inversly proportional to $\pi$
  
• C. Zero only if $\theta ={180}^{\circ }$
  
• D. Zero for all values of $\theta$
  

Answer 1

The correct option is D Zero for all values of $\theta$



Let $i_1$ and $i_2$ be the current flowing in arcs $\text{ACB}$ and $\text{ADB}$ respectively.

As the wire is uniform, its cross sectional area is constant throughout. So, the resistance depends only upon the length.

$R\propto L$

$\frac{R_1}{R_2}=\frac{R_{\text{ACB}}}{R_{\text{ADB}}}=\frac{L_{\text{ACB}}}{L_{\text{ADB}}}$

$L_{\text{ACB}}$ and $L_{\text{ADB}}$ represents the length of the arc.

$L_{\text{ACB}}=\text{radius}\times \left(\theta \right)$
$L_{ADB}=\text{radius}\times (2\pi -\theta )$

$\frac{L_{\text{ACB}}}{L_{\text{ADB}}}=\frac{\theta }{2\pi -\theta }$

As the battery is connected across $\text{AB}$ we can think $\text{ACB, ADB}$ are in parallel connection.

So potential difference across $\text{ACB}$ and $\text{ADB}$ is same.

$i_1,R_1\to$ current and resistance across $\text{ACB}$

$i_2,R_2\to$ current and resistance across $\text{ADB}$

Using Ohm's law we can write,

So, $i_1{R}_1=i_2{R}_2$

$\frac{i_1}{i_2}=\frac{R_2}{R_1}=\frac{R_{\text{ADB}}}{R_{\text{ACB}}}=\frac{L_{\text{ADB}}}{L_{\text{ACB}}}$

$=$ $\frac{2\pi -\theta }{\theta }$

Now,
$B_1\to$magnetic filed due to $\text{ACB}$.

$B_1=\frac{{\mu }_0{i}_1}{2\pi }\times \frac{\theta }{2r}$

$B_2\to$magnetic filed due to $\text{ADB}$.

$B_2=\frac{{\mu }_0{i}_1}{2\pi }\times \frac{(2\pi -\theta )}{2r}$

Now,$\frac{B_1}{B_2}=\frac{\frac{{\mu }_0{i}_1}{2\pi }\frac{\theta }{2r}}{\frac{{\mu }_0{i}_2}{2\pi }\frac{(2\pi -\theta )}{2r}}$

$\frac{i_1}{i_2}\times \frac{\theta }{(2\pi -\theta )}$

$\frac{2\pi -\theta }{\theta }\times \frac{\theta }{2\pi -\theta }=1$

That means, $B_1=B_2$

Both the magnetic fields are equal in magnitude but opposite in direction.

Why this Question ? Key point : Although the current in the two arcs are different, still the magnetic field have the same magnitude but opposite directions thereby producing zero net magnetic field.