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Question
A square loop of side 'a' carrying a current I2 is kept at distance x from an infinitely long straight wire carrying a current I1 as shown in the figure. Obtain the expression for the resultant force acting on the loop.
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Solution
According to the right-hand screw rule, the magnetic field will be into the plane across the loop Force on length AD
F = Bil
`F_1 = (mu_0I_1I_2a)/(2pix)`
Force on length BC
F = Bil
`F_2 = (mu_0I_1I_2a)/(2pi(x + a))`
Force on AB and CD will be equal and opposite . Hence, they'll cancel out. Force on the loop
`F_"Net" = F_1 - F_2`
= `(mu_0I_1I_2a)/(2pi)[1/x - 1/((x + a))]`
`F_("Net") = (mu_0I_1I_2a)/(2pi)[(x + a - x)/((x + a)x)] = (mu_0I_1I_2a^2)/(2pi(x + a)x)`
`F_("Net") = (mu_0I_1I_2a^2)/(2pix(x + a))` (Towards left)
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