Question

A wire loop formed by joining two semi-circular wires of radii $R^1$ arid $R^2$ carries a current as shown in the adjoining diagram. The magnetic induction at the centre O is :

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

Answer 1

The correct option is D $\frac{{\mu }_{0}I\pi }{4\pi }(\frac{1}{R_1}+\frac{1}{R_2})$

Answer is D.
Circular wire produces magnetic field inside the circle and outside the circle. Magnetic field around a circular wire is calculated by the formula;$B=\frac{{\mu }_{0}I}{2r}where,{\mu }_0=4\pi \times {10}^{-7}N/A-m$.
since we have semicircle of radius R1 below, we put 1/2 in front of our formula.
The formula becomes $\frac{{\mu }_{0}I\pi }{4\pi R_1}$.
Similarly, we have semicircle of radius R2 above, we put 1/2 in front of our formula.
The formula becomes $\frac{{\mu }_{0}I\pi }{4\pi R_2}$.
Therefore, $\frac{{\mu }_{0}I\pi }{4\pi R_1}$ + $\frac{{\mu }_{0}I\pi }{4\pi R_2}$
That is, the magnetic induction at the centre O is given as $\frac{{\mu }_{0}I\pi }{4\pi }(\frac{1}{R_1}+\frac{1}{R_2})$.