Question 37 of 276

Two insulated circular loop $A$ and $B$ radius $a$ carrying a current of $I$ in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be:

A$\dfrac{\sqrt{2}{\mu }_{0}I}{a}$B$\dfrac{{\mu }_{0}I}{2a}$C$\dfrac{{\mu }_{0}I}{\sqrt{2}a}$D$\dfrac{2{\mu }_{0}I}{a}$

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Answer: (C) $\dfrac{{\mu }_{0}I}{\sqrt{2}a}$

Right hand thumb rule states that, “If you imagine holding a current carrying wire in your right-hand with your thumb pointing towards the direction of electric current flow then the direction in which your fingers curl, gives the direction of lines of force of the magnetic field”. In this case, direciton of magnetic field at the centre for both wires is as shown below.

Magnetic field due to a circular wire is given by, ${B}_{A}={B}_{B}=\dfrac{{\mu }_{0}I}{2a}$
As both magnetic field lines at centre are perpendicular, therefore
$∵{B}_{net}=\sqrt{{\left({B}_{A}\right)}^{2}+{\left({B}_{B}\right)}^{2}}=\dfrac{\sqrt{2}{\mu }_{0}I}{2a}=\dfrac{{\mu }_{0}I}{\sqrt{2}a}$