Question

Wires 1 and 2 carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. The force on a small element ‘dl’ of wire 2 at a distance ‘n’ from the wire 1, as shown in figure, due to the magnetic field of wire 1 is

• A. $\frac{{\mu }_0}{2\pi \gamma }{i}_1{i}_{2}dltan\theta$
• B. $\frac{{\mu }_0}{2\pi \gamma }{i}_1{i}_{2}dlsin\theta$
• C. $\frac{{\mu }_0}{2\pi \gamma }{i}_1{i}_{2}dlcos\theta$
• D. $\frac{{\mu }_0}{2\pi \gamma }{i}_1{i}_{2}dlsin\theta$

Answer 1

The correct option is C $\frac{{\mu }_0}{2\pi \gamma }{i}_1{i}_{2}dlcos\theta$

$Now,\frac{F}{1}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}\therefore F=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_2}{p}.dlcos\theta$