Question

The magnetic induction at the center $\text{O}$ in the figure is,

• A. $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}-\frac{1}{R_2})$
• B. $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}+\frac{1}{R_2})$
• C. $\frac{{\mu }_{0}I}{4}(R_1-R_2)$
• D. $\frac{{\mu }_{0}I}{4}(R_1+R_2)$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{4}(\frac{1}{R_1}-\frac{1}{R_2})$

Magnetic field due to the two straight wire segments at $\text{O}$ is zero because point $\text{O}$ is lying on the axis of wire. So magnetic field at point $\text{O}$ will be produced due to semicircular segments only.

Thus, the magnetic field due to the inner semicircle of radius $R_1$ is given by,

$B_1=\frac{{\mu }_{0}I}{4R_1}$

Now using right- hand thumb rule for the direction of magnetic field

$\therefore B_1=\frac{{\mu }_{0}I}{4R_1}⨂$

$⨂$ indicates the direction of magnetic field perpendicularly inwards to the plane of paper.

Similarly, the magnetic field due to the outer semicircle of radius $R_2$ is given by,

$B_2=\frac{{\mu }_{0}I}{4R_2}$

$\therefore B_2=\frac{{\mu }_{0}I}{4R_2}⨀$

So, the net magnetic field at point $\text{O}$ will be

$\left|\begin{array}{c}\overrightarrow{B_{net}}\end{array}\right|=\left|\begin{array}{c}\overrightarrow{B_1}\end{array}\right|-\left|\begin{array}{c}\overrightarrow{B_2}\end{array}\right|$

(considering inward direction to the plane of paper is positive)

$B_{net}=[\frac{{\mu }_{o}I}{4R_1}-\frac{{\mu }_{o}I}{4R_2}]$

$\therefore B_{net}=\frac{{\mu }_{o}I}{4}[\frac{1}{R_1}-\frac{1}{R_2}]$

Thus, the direction of the resultant field will be perpendicular and into the plane of page.