A long,straight wire is fixed horizontally and carries a current of 50.0 A.A second wire having linear mass density $1.0 \times 10^{- 4} k g m^{- 1}$ is placed parallel to and directly above this wire at separation of 5.0 mm.What current should this second wire carry such that the magnetic repulsion can balance its weight ?

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Solution

According to question $\frac{μ_{0} i_{1} i_{2}}{2 π d} = \frac{d F}{d l}$

$\frac{2 \times 10^{- 7} \times 50 \times i_{2}}{5 \times 10^{- 3}} = 1 \times 10^{- 4} \times 9.8$

$i_{2} = \frac{9.8 \times 10^{- 7}}{20 \times 10^{- 7}} A$

0.49 A in opposite direction.

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