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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 - Physics

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Answer in Brief

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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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)

Concept: Torque on Current Loop, Magnetic Dipole - Torque on a Rectangular Current Loop in a Uniform Magnetic Field
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