First slide

Force on a current carrying wire placed in a magnetic field

Question

A square loop ABCD of side length 2m is placed mean a very long straight conductor PQ carrying a current of 10 A as shown in figure. The square loop is carrying a current of 10 A and the conductor PQ lies in the plane of the loop ABCD. Then the magnetic force exerted by the square loop on the straight conductor is

Moderate

Solution

Force experienced by the square loop
$= F_{AB} - F_{CD} = i . a . \frac{μ_{0} i}{2 πa} - i . a . \frac{μ_{0} i}{2 π . 2 a} = \frac{μ_{0} i^{2}}{4 π} = 10^{- 7} \times {(10)}^{2} N$
$= 10^{- 5} N$
By Newton's third law of motion, force exerted by the square loop on $PQ = 10^{- 5} N$

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