Question

A wire loop formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current as shown in the adjoining diagram. The magnetic induction at the centre $O$ is:

A. $\dfrac{\mu_{0}I}{4R_{1}}$
B. $\dfrac{\mu_{0}I}{4R_{2}}$
C. $\dfrac{\mu_{0}I}{4}\left(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}}\right)$
D. $\dfrac{\mu_{0}I}{4}\left(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}\right)$

Answer 1

Hint: We know that a current carrying conductor can produce magnetic fields around itself. And we also know that current can be produced by a varying magnetic field. The current produced is called induced current and its direction is given by right hand thumb rule.

Formula used:
${{B}_{centre}}=\dfrac{{{\mu }_{0}}I}{2 R}$

Complete step-by-step answer:
We know that a magnetic field induction refers to the production of electromotive force across a current carrying conductor in a changing magnetic field. Or, in simple terms, when current is passed through a coil, a magnetic field is created in the coil. This magnetic field restricts the current to flow through the coil. From Biot-Savart law, we know that magnetic field induction at the centre of a current carrying coil is given by
${{B}_{centre}}=\dfrac{{{\mu }_{0}}I}{2 R}$
where
${{B}_{centre}}$ is the magnetic field induction at the centre of a coil
${{\mu }_{0}}$ is the magnetic constant or permeability in free space
$I$ is the current flowing through the coil
$R$ is the radius of the coil
Thus, clearly, the magnetic field is inversely proportional to the radius of the coil. Here, we have two radius, then the net magnetic field is the sum of the magnetic field due to the two wires of different radii, since we have only a part of the wire i.e. half, we can take
${{B}_{centre}}=\dfrac{{{\mu }_{0}}I}{4 R}$

Then, $B=B_1+B_2=\dfrac{{{\mu }_{0}}I}{4 R_1}+\dfrac{{{\mu }_{0}}I}{4 R_2}$
$\implies B=\dfrac{{{\mu }_{0}}I}{4}\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$

Then from the right hand thumb rule, we see that when winding the fingers along the direction of current, we see that the thumb points into the paper. Thus the direction of the induced magnetic field is into the paper with a magnitude of $B=\dfrac{{{\mu}_{0}}I}{4}\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)$
Hence the correct answer s option D. $\dfrac{\mu_{0}I}{4}\left(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}\right)$

So, the correct answer is “Option D”.

Note: That direction of current and the direction of magnetic field are interrelated; when one is given the other can be found using the right hand thumb rule. Also note that the current carrying conductor produces magnetic field and change in magnetic flux produces current.