The figure given below shows a square-shaped loop MNOP of side \(25\text{ cm}\) placed horizontally in a uniform magnetic field \(\vec{B}\) directed vertically downward. The position of the loop at \(t = 0\text{ s}\) is as shown in figure. The loop is pulled with a constant velocity of \(25\text{ cm/s}\) till it goes out of the field.
(a) What will be the direction of the induced current in the loop as it goes out of the field ? For how long would the current in the loop persist ?
(b) Plot graphs showing the variation of magnetic flux and magnitude of induced emf as a function of time.
Show Hint
Whenever a closed loop moves through a localized uniform magnetic field of width \(L\), the time taken to enter or leave depends solely on the size of the loop itself (\(\Delta t = \text{side}/v\)), whereas the duration of zero induction inside depends on the gap width (\(t_{\text{flat}} = (L - \text{side})/v\)). This structural intuition makes sketching verification values instantly straightforward!
Step 1: Read the data and the figure:
Side of the square loop: \(l = 25\text{ cm} = 0.25\text{ m}\).
Speed of the loop: \(v = 25\text{ cm/s} = 0.25\text{ m/s}\).
The field region is \(1\text{ m}\) long. At \(t = 0\) the whole loop lies inside the field, with side MP at the left edge of the field.
So the loop covers the first \(25\text{ cm}\) of the \(1\text{ m}\) region, and the front side NO is \(75\text{ cm}\) from the right edge.
Step 2: Find the time interval of each stage:
The loop stays fully inside until side NO reaches the right edge of the field.
Distance to travel: \(1\text{ m} - 0.25\text{ m} = 0.75\text{ m}\).
\[ t_1 = \frac{0.75\text{ m}}{0.25\text{ m/s}} = 3\text{ s} \]
The loop is completely outside when side MP also crosses the right edge, after a further \(0.25\text{ m}\):
\[ t_2 = \frac{1.00\text{ m}}{0.25\text{ m/s}} = 4\text{ s} \]
Step 3: Direction of the induced current (part a):
The field is vertically downward. Seen from above, it points into the plane of the figure (the crosses).
For \(3\text{ s} < t < 4\text{ s}\) the area inside the field decreases, so the inward flux decreases.
By Lenz's law the induced current opposes this decrease. It must produce an inward field inside the loop.
An inward field inside a loop is produced by a clockwise current.
The current flows along \(M \rightarrow N \rightarrow O \rightarrow P \rightarrow M\), that is clockwise as seen from above.
Step 4: Time for which the current persists (part a):
A current flows only while the flux changes. The flux changes only while the loop leaves the field.
\[ \Delta t = t_2 - t_1 = 4\text{ s} - 3\text{ s} = 1\text{ s} \]
Check: \(\Delta t = \dfrac{l}{v} = \dfrac{0.25}{0.25} = 1\text{ s}\).
Step 5: Magnetic flux against time (part b):
The flux through the loop is \(\Phi = B \times (\text{area inside the field})\). The maximum value is \(\Phi_0 = Bl^2 = B(0.25)^2 = 0.0625\,B\) (in Wb, with \(B\) in tesla).
\(0 \le t \le 3\text{ s}\): loop fully inside, so \(\Phi = \Phi_0\), a horizontal straight line.
\(3\text{ s} \le t \le 4\text{ s}\): \(\Phi = Bl\,[\,l - v(t-3)\,]\). This is a straight line falling from \(\Phi_0\) to \(0\).
For \(t > 4\text{ s}\): \(\Phi = 0\) (on the time axis).
Step 6: Magnitude of induced emf against time (part b):
\(0 \le t < 3\text{ s}\): \(\dfrac{d\Phi}{dt} = 0\), so \(|\varepsilon| = 0\).
\(3\text{ s} < t < 4\text{ s}\): \(|\varepsilon| = \left|\dfrac{d\Phi}{dt}\right| = Blv = B(0.25)(0.25) = 0.0625\,B\) volt, a constant value (a rectangular pulse).
For \(t > 4\text{ s}\): \(|\varepsilon| = 0\).
Draw the \(\Phi\)-\(t\) graph as a flat top until \(3\text{ s}\), then a straight fall to zero at \(4\text{ s}\). Draw the \(|\varepsilon|\)-\(t\) graph as zero until \(3\text{ s}\), a flat pulse of height \(Blv\) from \(3\text{ s}\) to \(4\text{ s}\), then zero.
Final Answer:
The induced current is clockwise (seen from above) and it persists for \(1\text{ s}\) (from \(t = 3\text{ s}\) to \(t = 4\text{ s}\)). The flux is constant at \(Bl^2\) up to \(3\text{ s}\), then falls linearly to zero at \(4\text{ s}\). The emf is zero up to \(3\text{ s}\), equals \(Blv\) from \(3\text{ s}\) to \(4\text{ s}\), and is zero afterwards.
