question_answer

In the given figure, the wires AB, CD and EF have identical resistances. The separation between the neighbouring wire is 1.0 cm.

The wires AE and BF have negligible resistance and the ammeter reads 15 A.

Calculate the magnetic force per unit length of AB and CD.

Answer:

Force per unit length of AB \[=5\times {{10}^{-6}}N\] Force per unit length of CD = zero As the wires AB, CD and EF have identical resistance, they carry same current \[=\frac{15}{3}=5A\] \[{{I}_{1}}={{I}_{2}}={{I}_{3}}=5A\] \[d=100\,\,cm\] Magnetic force per unit length on wire AB                         \[F=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{I}_{1}}{{I}_{2}}}{d}={{10}^{-7}}\times \frac{2\times 5\times 5}{1}\]                         \[=5\times {{10}^{-6}}N\] As the wires AB and CD has current is same direction, there is an attractive force between them. The wire CD experiences, attractive magnetic forces due to AB and EF which are of equal magnitude and opposite in direction. Hence, net magnetic force on CD = 0.