question_answer

Two concentric coils each of radius equal to \[2\pi cm\]are placed right angles to each other. If 3 A and 4 A are the currents flowing through the two coils respectively. The magnetic induction (in Wb\[{{m}^{-2}}\]) at the centre of the coils will be

A)  \[{{10}^{-5}}\]            

B)  \[7\times {{10}^{-5}}\]

C)  \[12\times {{10}^{-5}}\]        

D)  \[5\times {{10}^{-5}}\]

Correct Answer: D

Solution :

: The situation is shown in the figure. For coil 1, Radius,  \[{{R}_{1}}=2\pi \,cm\,=2\pi \times {{10}^{-2}}m\] Current, \[{{I}_{1}}=3A\] For coil 2, Radius, \[{{R}_{2}}=2\pi cm=2\pi \times {{10}^{-2}}m\] Current, \[{{I}_{2}}=4A\] The magnetic induction at the centre O due to the current \[{{I}_{1}}\] in coil 1 is \[{{B}_{1}}=\frac{{{\mu }_{0}}{{I}_{1}}}{2{{R}_{1}}}\] It acts in vertically upwards direction. The magnetic induction at the centre O due to the current \[{{I}_{2}}\] in coil 2 is \[{{B}_{2}}=\frac{{{\mu }_{0}}{{I}_{2}}}{2{{R}_{2}}}\] It acts in the horizontal direction. As \[{{B}_{1}}\] and \[{{B}_{2}}\] are perpendicular to each other, so the net magnetic field at the centre O is \[B=\sqrt{B_{1}^{2}+B_{2}^{2}}\] \[=\sqrt{{{\left( \frac{{{\mu }_{0}}{{I}_{1}}}{2{{R}_{1}}} \right)}^{2}}+{{\left( \frac{{{\mu }_{0}}{{I}_{2}}}{2{{R}_{2}}} \right)}^{2}}}\] Substituting the given values, we get \[B=\sqrt{{{\left( \frac{{{\mu }_{0}}(3A)}{2(2\pi \times {{10}^{-2}}m} \right)}^{2}}+{{\left( \frac{{{\mu }_{0}}(5A)}{2(2\pi \times {{10}^{-2}}m)} \right)}^{2}}}\] \[=\frac{{{\mu }_{0}}\sqrt{{{(3A)}^{2}}+{{(4A)}^{2}}}}{2(2\pi \times {{10}^{-2}}m)}\] \[=\frac{(4\pi \times {{10}^{-7}}Wb\,{{m}^{-1}}{{A}^{-1}})(5A)}{(4\pi \times {{10}^{-2}}m)}\] \[=5\times {{10}^{-5}}Wb\,{{m}^{-2}}\]