Magnetic field

The motion of the loop can be broken into three phases.

1) Loop entering into the magnetic field.

2) Loop is completely inside the magnetic field.

3) Loop is coming out of the magnetic field.

In the first case, let the time be $t$ (after the loop starts entering in the magnetic field) the speed of the loop becomes $v$ and the loop has penetrated into the region of the magnetic field to a distance $x$ . The magnetic flux $\phi$ passing through the loop will be,

$\phi = B (ax)$ and emf $\epsilon = \dfrac{d\phi}{dt} = Ba \dfrac{dx}{dt} = Bav$ .

This gives the current in the loop, $i = \dfrac{\epsilon}{R} = \dfrac{Bav}{R}.$

The loop will experience a force $F = BiL = \dfrac{B^2a^2v}{R}$ . According to Lenz's law , this force will decelerate the loop by an acceleration $a = -v\dfrac{dv}{dx} = \dfrac{B^2a^2v}{mR}$ .

$\int_{0}^{\Delta v} \, dv = - \int_0^a {\frac{B^2a^2}{mR}} \,dx \\ \Delta v = - \frac{B^2a^3}{mR} .$

Therefore, during the entry of the loop, its speed decreases by $\dfrac{B^2a^3}{mR}$ . An equal amount of change in speed will occur when the loop comes out of the magnetic field.

When the loop is completely inside the magnetic field then the flux will not change and hence no EMF or force will be generated on the loop.

Therefore, the loop will come out of the magnetic field at a speed, $v = v_0 -2\Delta v = \frac{3B^2a^3}{mR}-\frac{2B^2a^3}{mR} = \frac{B^2a^3}{mR} = \frac{v_0}{3} .$

The loop loses its kinetic energy in the form of heat, therefore, heat loss is equal to the loss in kinetic energy.

Heat loss = $\dfrac{1}{2} m \left({v_0}^2 - \left(\dfrac{v_0}{3}\right)^2\right) = \boxed{\dfrac{4}{9} m v_0^2}.$