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

$$B_2=\frac{{\mu }_o}{4\pi }\frac{2I}{r}$$

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

$$B_2=\frac{{\mu }_o}{4\pi }\frac{2\times 3I}{r}$$

$$B_2=\frac{{\mu }_o}{4\pi }(\frac{6I}{r})$$

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

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

$$B_3=\frac{{\mu }_o}{4\pi }\frac{2\times 4I}{r+2r}$$

$$B_3=\frac{{\mu }_o}{4\pi }(\frac{8I}{3r})$$

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

$$\text{ Net Magnetic field at point A }$$

$$B=B_2-B_3$$

$$=\frac{{\mu }_o}{4\pi }(\frac{3I}{r})-\frac{{\mu }_o}{4\pi }(\frac{8I}{3r})$$

$$=\frac{{\mu }_o}{4\pi }(\frac{I}{3r})$$

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

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

$$F_{21}=\frac{{\mu }_o}{4\pi }\frac{2I_2{I}_1}{r}$$

$$=\frac{{\mu }_o}{4\pi }\frac{2\times 3I\times I}{r}=\frac{{\mu }_o}{4\pi }\frac{6I^2}{r}$$

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

$$F_{23}=\frac{{\mu }_o}{4\pi }\frac{2I_2{I}_3}{r}$$

$$=\frac{{\mu }_o}{4\pi }\frac{2\times 3I\times 4I}{2r}=\frac{{\mu }_o}{4\pi }\frac{12I^2}{r}$$

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

$$=\frac{{\mu }_o}{4\pi }\frac{12I^2}{r}-\frac{{\mu }_o}{4\pi }\frac{6I^2}{r}$$

$$=\frac{{\mu }_o}{4\pi }\frac{6I^2}{r}$$