Two parallel straight wires X and Y separated by a distance 5 cm in air carrycurrent of 10 A and 5 A respectively in opposite direction as shown in thefigure. Calculate magnitude and direction of the force on a 20 cm length ofthe wire Y.
Question
Two parallel straight wires X and Y separated by a distance 5 cm in air carrycurrent of 10 A and 5 A respectively in opposite direction as shown in thefigure. Calculate magnitude and direction of the force on a 20 cm length ofthe wire Y.
Solution
The force between two parallel wires carrying current is given by Ampere's force law:
F = (μ0 * I1 * I2 * L) / (2πr)
where:
- F is the force between the wires,
- μ0 is the permeability of free space (4π x 10^-7 T m/A),
- I1 and I2 are the currents in the wires,
- L is the length of the wire, and
- r is the distance between the wires.
Substituting the given values:
F = (4π x 10^-7 T m/A * 10 A * 5 A * 0.2 m) / (2π * 0.05 m)
Solving the above expression gives:
F = 8 x 10^-5 N
The direction of the force is attractive because the currents are in opposite directions. Therefore, wire Y is attracted towards wire X.
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