Magnetic force on a loop

Consider a vertical strip inside the loop of infinitesimal horizontal width $dx$ and vertical length $a$ which lies at a distance $x$ from the infinitely long current carrying wire. The magnetic flux associated with this infinitesimal area will be

$d \Phi = B \,dS = \dfrac{\mu_0 I}{2\pi x} \cdot a \,dx$

Total magnetic flux through the loop would thus be

$\Phi_B = \int_a^{2a} \dfrac{\mu_0 I}{2\pi x} \cdot a \,dx = \dfrac{\mu_0 I a}{2\pi} \ln(2) = \dfrac{\mu_0 I_0 t a}{2\pi} \ln(2)$

Using Faraday's law's, the emf generated in the loop will be

$\mathcal{E} = \left| \dfrac{d \Phi_B}{dt} \right| = \dfrac{\mu_0 I_0 a}{2\pi} \ln(2)$

and the current associated with the loop will be

$i = \dfrac{\mathcal{E}}{R} = \dfrac{\mu_0 I_0 a}{2\pi R} \ln(2)$

We see that the direction of magnetic field due to the current carrying wire is into the plane of the loop and the current is increasing linearly with time in the wire. Therefore, the flux associated with the loop is inward and increasing. According to Lenz's law, the direction of current should be thus anticlockwise so as to oppose the change in magnetic flux.

Since for every infinitesimal current carrying element experiencing a force on one horizontal side of the loop there is another infinitesimal current carrying element experiencing exactly the same amount of force but in opposite direction, the net force for the horizontal sides of the loop comes out to zero.

For the vertical sides, every infinitesimal element on one particular side experiences the same force due to same magnetic field and the same current flowing through it. But the magnetic field strength decreases with distance. Hence there is a net force acting in the horizontal direction which is away from the infinitely long current carrying wire and this force is given by

$F = i a \left( \dfrac{\mu_0 I}{2\pi a} - \dfrac{\mu_0 I}{2\pi (2a)} \right) = \dfrac{\mu_0 I_0 a}{2\pi R} \ln(2) \left( \dfrac{\mu_0 I_0 t}{2\pi (2)} \right) = \boxed{1.386}$