question_answer

Two long parallel wires P and Q are held perpendicular to the plane of the paper at a separation of 5 m. If P and Q carry currents of 2.5 A and 5 A respectively in the same direction, then the magnetic field at a point midway between P and Q is

A) \[\frac{{{\mu }_{0}}}{\pi }\]                 

B)        \[\sqrt{3}\frac{{{\mu }_{0}}}{\pi }\]

C) \[\frac{{{\mu }_{0}}}{2\pi }\]               

D)        \[\frac{3{{\mu }_{0}}}{2\pi }\] 

Correct Answer: C

Solution :

When the current flows in both wires in the same direction then magnetic field at half way due to the wire P, \[{{\vec{B}}_{p}}=\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi \frac{5}{2}}=\frac{{{\mu }_{0}}{{I}_{1}}}{\pi \cdot 5}=\frac{{{\mu }_{0}}}{2\pi }\] \[(where\,\,{{I}_{1}}=5amp)\] The direction of \[{{\vec{B}}_{p}}\] is downward Q Magnetic field at halfway due to wire Q \[{{\vec{B}}_{Q}}=\frac{{{\mu }_{0}}{{I}_{2}}}{2\pi \frac{5}{2}}=\frac{{{\mu }_{0}}}{\pi }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[upward\,\,\odot ]\] \[[where\,\,{{I}_{2}}=5\,\,amp.]\] Net magnetic field at halfway \[\vec{B}={{\vec{B}}_{p}}+{{\vec{B}}_{Q}}=\,\,-\frac{{{\mu }_{0}}}{2\pi }+\frac{{{\mu }_{0}}}{\pi }=\frac{{{\mu }_{0}}}{2\pi }\,\,(upward)\] Hence, net magnetic field at midpoint \[=\frac{{{\mu }_{0}}}{2\pi }\]