Question

The figure shows a part of an electric circuit. The wires AB,CD and EF are long and have identical resistances. The separation between the neighboring wires is 1.0cm . The wires EA and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length on AB and CD.

Answer 1

A current of $I=30A$ passes through ammeters which gets distributed in the three wires As resistance of each wire are equal, thin current gets distributed to the three wires. So, current in each $AB,CD,EF=\frac{30}{3}=10A$

Magnetic force per length on $AB$ will be :-$[I_{AB}=I_{CD}=I_{EF}=10A$

$B_1=\frac{\mu I_{AB}{I}_{CD}}{2\pi r}+\frac{{\mu }_0{I}_{AB}{I}_{EF}}{2\pi \left(2r\right)}$

$=\frac{3}{4}\frac{\mu o{I}^2}{\pi r}=\frac{3}{4}\times \frac{4\pi \times {10}^{-7}\times {10}^2}{\pi \times {10}^{-2}}=3\times {10}^{-3}T=3mN/m$

and that on $CD$,

Notice that, the magnetic force on $CD$ by $AB$ and $EF$ are equal but in opposite directions, no force on $CD=O$