(i) State Lenz's law and explain how this law is a consequence of conservation of energy principle.
(ii) A square shaped loop of side \(\dfrac{\ell}{2}\) is initially lying outside a region of uniform magnetic field \(\vec{B}\) as shown in the figure. The loop is moved towards right with a constant velocity \(\vec{v}\) till it goes out of the region of magnetic field.
(I) What will be the directions of induced current when the loop enters the field and when it leaves the field ?
(II) Draw the plots showing the variation of magnetic flux \(\phi\) linked with the loop with time \(t\) and variation of induced emf \(E\) with time \(t\). Mark the relevant values of \(E\), \(\phi\) and \(t\) on the graphs.
Show Hint
Induced current opposes the flux change; flux is Ba^2 inside, emf is Bav only while entering or leaving.
Step 1: Lenz's law and energy conservation (i):
Lenz's law: the direction of the induced emf (and current) is such that it opposes the change in magnetic flux that produces it.
Suppose it did the opposite. Then pushing a magnet towards a coil would induce a current that pulls the magnet further in. The magnet would speed up, and the current would grow, with no one doing work.
This would create energy from nothing, and so it breaks the law of conservation of energy. Because the induced current opposes the motion, we must do mechanical work against it. That work appears as electrical energy in the loop. So Lenz's law is a result of energy conservation.
Step 2: Setting up the motion (ii):
From the figure, the field region is of width \(\ell\), the field points into the page, the loop has side \(a = \ell/2\) and it starts a distance \(\ell\) before the field. Speed is \(v\).
Times: the loop starts entering at \(t_1 = \ell/v\). It is fully inside at \(t_2 = \dfrac{3\ell}{2v}\) (it moves a distance \(\ell/2\)). It starts to leave at \(t_3 = \dfrac{2\ell}{v}\) and is fully out at \(t_4 = \dfrac{5\ell}{2v}\).
Step 3: Directions of induced current (ii)(I):
While entering, the flux into the page increases. The loop opposes this by producing a field out of the page inside it, so the induced current is anticlockwise.
While leaving, the flux into the page decreases. The loop opposes this by producing a field into the page, so the induced current is clockwise.
Step 4: Flux and emf graphs (ii)(II):
Maximum flux when the loop is fully inside: \(\phi_m = B a^2 = \dfrac{B\ell^2}{4}\). The emf magnitude while it enters or leaves is \(E = B a v = \dfrac{B\ell v}{2}\).
Flux graph (\(\phi\) against \(t\)): zero from \(0\) to \(\ell/v\); a straight line rising to \(\dfrac{B\ell^2}{4}\) at \(\dfrac{3\ell}{2v}\); flat at \(\dfrac{B\ell^2}{4}\) up to \(\dfrac{2\ell}{v}\); a straight line falling to zero at \(\dfrac{5\ell}{2v}\); zero afterwards.
emf graph (\(E\) against \(t\)): zero from \(0\) to \(\ell/v\); a constant value \(-\dfrac{B\ell v}{2}\) from \(\ell/v\) to \(\dfrac{3\ell}{2v}\); zero from \(\dfrac{3\ell}{2v}\) to \(\dfrac{2\ell}{v}\); a constant value \(+\dfrac{B\ell v}{2}\) from \(\dfrac{2\ell}{v}\) to \(\dfrac{5\ell}{2v}\); zero afterwards. The signs are opposite because the current reverses (taking flux into the page as positive, \(E = -d\phi/dt\)).
Final Answer:
Entering: anticlockwise current. Leaving: clockwise current. Maximum flux is \(\dfrac{B\ell^2}{4}\) and the emf magnitude is \(\dfrac{B\ell v}{2}\) in both the entering and leaving intervals.\[ \boxed{\phi_{max} = \dfrac{B\ell^2}{4}, \quad |E| = \dfrac{B\ell v}{2}} \]
