A rectangular loop of sides 25 cm and 10 cm carrying current of 15 A is placed with its longer side parallel to a long straight conductor 2.0 cm apart carrying a current of 25 A. What is the new force on the loop?
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Solution
When the loop is placed with the longest side parallel to the long straight wire, then the shortest sides of the loop at the same distance from the wire and force on them are equal but in opposite direction therefor net force on two short side are zero.
Now longest side which near the wire, force on it is given by
F1=B1IloopL
where
B is the magnetic field due to wire L is the length of the side of the loop
B=μ0I2πr
where r is the distance of the sides of the loop
Let direction of the current in wire is in upward direction.
F1=μ0IwireIloop2πr
F1=4π×10−7×25×15×252π×2
F1=9.37×10−4N away from the wire.
Force on the longest side which is away from the wire