Question

1. Figure below shows the front view of two straight wires you have in equilibrium, each carrying current I in opposite direction, one into and the other out of the page. Calculate the magnitude of current I.​

Answer 1

Answer

Solution

the magnetic field at O due to straight wire carrying current is

B→=μ04πIa[sin30∘+sin30∘]kˆ

=μ04πIa[12+12]kˆ=μ04πIakˆ

Total magnetic field at O due to current through all the straight parallel wires carrying currents

B→=B1kˆ+B2(−kˆ)+B3(kˆ)+B4(−kˆ)+...

B=B1−B2+B3−B4+...

=μ04πIa[1−12+13−14+...]

=μ04πIaloge(1+1)=μ04πIaloge2

Here, cos30∘=ar or a=rcos30∘=r3–√/2

∴B=μ04πI(r3–√/2)loge2=μ02πIr3–√loge2

μ0I2πloge2r3–√  is the answer..