Question

A battery is connected between two points A and B on the circumference of a uniform conducting ring of radius $r$ and resistance $R$. One of the arcs AB subtends an angle $\theta$ at the centre. The value of the magnetic induction at the centre due to the current in the ring is :

• A. proportional to (${180}^o$ - $\theta$)
• B. inversely proportional to $r$
• C. zero only if $\theta$ $=$ ${180}^o$
• D. zero for all values of $\theta$

Answer 1

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

Direction of current $i_1$ and $i_2$ in two parts of the circuit are different.

Since the battery voltage is same for both parts, $i_1{R}_1=i_1{R}_2$ where $R_1$ and $R_2$ are resistances of the arcs respectively.

But,

$R_1=\frac{\rho l_1}{A}$

$R_1=\frac{\rho l_1}{A}$

$\therefore i_1{R}_1=i_1{R}_2$

$\Rightarrow i_1\frac{\rho l_1}{A}=i_2\frac{\rho l_2}{A}$

$\Rightarrow i_1{l}_1=i_2{l}_2$

Where A is the cross sectional area of the wire. $\rho$ is the resistivity of the wire.

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

$B_2=\frac{{\mu }_0{i}_2}{2\pi r}{l}_2$

The direction of the fields are opposite to each other.

$B=B_2-B_1=\frac{{\mu }_0{i}_2}{2\pi r}{l}_2-\frac{{\mu }_0{i}_1}{2\pi r}{l}_1=\frac{{\mu }_0}{2\pi r}(i_2{l}_2-i_1{l}_1)=0$