Physics - EMI/AC Question with Solution | TestHub

When the loop has entered a distance $x$ into the field, the magnetic flux $\Phi$ through the part of the loop inside the field region is given by integrating $B(x^{\prime })$ over the entered area:$\Phi ={\int }_0^{x}B(x^{\prime })(ad{x}^{\prime })={\int }_0^x(B_0+k{x}^{\prime })ad{x}^{\prime }=a{\left[B_0{x}^{\prime }+\frac{k{x}^{\prime 2}}{2}\right]}_0^x=a\left(B_{0}x+\frac{k{x}^2}{2}\right)$

The induced electromotive force (EMF) is

$\mathcal{E}=-\frac{d\Phi }{dt}$.

Since $x$ changes with time,

$x=vt$,

so $\frac{dx}{dt}=v$.

$\mathcal{E}=-a\left(B_0\frac{dx}{dt}+kx\frac{dx}{dt}\right)=-av(B_0+kx)$

The magnitude of the induced current is

$I=\frac{|\mathcal{E}|}{R}=\frac{av(B_0+kx)}{R}$.

To maintain constant velocity, an external force must balance the magnetic force on the loop. The magnetic force acts primarily on the leading edge of the loop (length $a$) which is at position $x$ and experiences the field $B(x)=B_0+kx$.$F_{magnetic}=IaB(x)=\left(\frac{av(B_0+kx)}{R}\right)a(B_0+kx)=\frac{a^{2}v(B_0+kx{)}^2}{R}$

Since this force opposes the motion (by Lenz's law), the external force required to maintain constant velocity must be equal in magnitude and opposite in direction.