#### Question

Answer in Brief

A square loop of side 'a' carrying a current I_{2} is kept at distance x from an infinitely long straight wire carrying a current I_{1} as shown in the figure. Obtain the expression for the resultant force acting on the loop.

#### 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)

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

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