Question

The figure shows three infinitely long straight parallel current carrying conductors. Find the
(i) magnitude and direction of the net magnetic field at point A lying on conductor 1,
(ii) magnetic force on conductor 2.

Answer 1

(i) Magnetic field due to a straight infinite long straight wire carrying current is given by

\[B_2 = \frac{\mu_o}{4\pi}\frac{2 I}{r}\]

where, I is the current flowing in the conductor.

Magnetic field (B₂​) at point A due to current carrying wire 2
Current in wire 2 is 3I

\[B_2 = \frac{\mu_o}{4\pi}\frac{2 \times 3 I}{r}\]

\[ B_2 = \frac{\mu_o}{4\pi}(\frac{6 I}{r})\]

\[\text { The direction of magnetic field at point A due to wire 2 is in inward direction } . \]

Magnetic field (B₃) at point A due to charge carrying wire 3
Current in wire 3 is 4I

\[B_3 = \frac{\mu_o}{4\pi}\frac{2 \times 4 I}{r + 2r}\]

\[ B_3 = \frac{\mu_o}{4\pi}(\frac{8 I}{3r})\]

\[\text { The direction of magnetic field at point A due to wire 3 is in the outward direction } . \]

\[\text { Net Magnetic field at point A }\]

\[B = B_2 - B_3 \]

\[ = \frac{\mu_o}{4\pi}(\frac{3 I}{r}) - \frac{\mu_o}{4\pi}(\frac{8 I}{3r})\]

\[ = \frac{\mu_o}{4\pi}(\frac{I}{3r}) \]

\[\text { The direction of net magnetic field at point A is in inward direction } . \]

(ii)
Magnitude of force on wire 2 due to wire 1 

\[F_{21} = \frac{\mu_o}{4\pi} \frac{2 I_2 I_1}{r}\]

\[ = \frac{\mu_o}{4\pi}\frac{2 \times 3I \times I}{r} = \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]

\[\text { Magnitude of force on wire 2 due to wire } 3\]

\[ F_{23} = \frac{\mu_o}{4\pi} \frac{2 I_2 I_3}{r}\]

\[ = \frac{\mu_o}{4\pi}\frac{2 \times 3I \times 4I}{2r} = \frac{\mu_o}{4\pi}\frac{12 I^2}{r}\]

\[\text { Net force on wire 2 } = F_{23} - F_{21} \]

\[ = \frac{\mu_o}{4\pi}\frac{12 I^2}{r} - \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]

\[ = \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]