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!
Concept: This problem involves Faraday's Law of Electromagnetic Induction and Lenz's Law, describing how changing magnetic flux produces an electromotive force (emf) and current in a closed loop. • Magnetic Flux (\(\Phi\)): The magnetic flux passing through a loop of area \(A\) completely inside a uniform magnetic field \(B\) directed perpendicular to its plane is given by: \[ \Phi = B \cdot A \] When the loop is partially outside, \(A\) represents the shaded region of the loop still submerged in the magnetic field. • Faraday's Law of Induction: The magnitude of the induced electromotive force (\(\varepsilon\)) is directly proportional to the time rate of change of magnetic flux through the circuit: \[ \varepsilon = -\frac{d\Phi}{dt} \] • Lenz's Law: The direction of the induced current is always such that it sets up an opposing magnetic field to counteract the change in original magnetic flux that created it.
Step 1: Determine the direction of the induced current as the loop exits the field. As the square loop \(MNOP\) is pulled horizontally to the right out of the uniform inward magnetic field (indicated by the \(\times\) symbols): • The effective surface area of the loop remaining inside the magnetic field decreases continuously over time. • Consequently, the inward magnetic flux passing through the interior area of the loop decreases. • According to Lenz's Law, the induced current must create an auxiliary magnetic field pointing into the plane of the page (\(\times\)) to reinforce the diminishing flux. • Applying the Right-Hand Grip Rule, a clockwise current produces a magnetic field directed into the page. Therefore, the direction of the induced current around the perimeter of the loop is clockwise (along the path \(M \rightarrow N \rightarrow O \rightarrow P \rightarrow M\)).
Step 2: Calculate the specific time intervals for the motion. Let us establish the exact timing of the loop's entry, transit, and exit phases. Let the side length of the square loop be \(l = 25\text{ cm} = 0.25\text{ m}\). The velocity at which it is pulled is \(v = 25\text{ cm/s} = 0.25\text{ m/s}\). The total horizontal extent of the uniform magnetic field region is \(L = 1\text{ m}\). • Initial Position (\(t = 0\)): The leading vertical edge \(NO\) of the loop is just aligned at the left boundary of the \(1\text{ m}\) wide field. • Completely inside the field (\(t_1\)): The loop becomes fully immersed when its trailing vertical edge \(MP\) travels a distance equal to its own width \(l = 0.25\text{ m}\): \[ t_1 = \frac{l}{v} = \frac{25\text{ cm}}{25\text{ cm/s}} = 1\text{ s} \] Between \(t = 0\text{ s}\) and \(t = 1\text{ s}\), the flux increases steadily until the loop is completely inside. • Reaching the right boundary (\(t_2\)): The leading vertical edge \(NO\) reaches the right edge of the magnetic field after traveling a full distance of \(L = 1\text{ m}\): \[ t_2 = \frac{L}{v} = \frac{1\text{ m}}{0.25\text{ m/s}} = 4\text{ s} \] Between \(t = 1\text{ s}\) and \(t = 4\text{ s}\), the loop remains fully submerged inside the uniform field. The flux stays constant, so the induced emf is zero during this phase. • Completely outside the field (\(t_3\)): The loop begins exiting at \(t = 4\text{ s}\). The trailing edge \(MP\) clears the right edge of the field after traveling an additional distance equal to the loop width \(l = 0.25\text{ m}\): \[ t_3 = \frac{L + l}{v} = \frac{1\text{ m} + 0.25\text{ m}}{0.25\text{ m/s}} = 4\text{ s} + 1\text{ s} = 5\text{ s} \]
Step 3: Determine the duration for which the current persists. An induced current only flows when there is a non-zero induced electromotive force, which requires a changing magnetic flux (\(\frac{d\Phi}{dt} \neq 0\)). As found in Step 2, the loop is exiting the field from \(t = 4\text{ s}\) to \(t = 5\text{ s}\). The duration (\(\Delta t\)) for which the loop undergoes this exit phase is: \[ \Delta t = t_3 - t_2 = 5\text{ s} - 4\text{ s} = 1\text{ s} \] Thus, the induced current during the exit phase persists for exactly \(1\text{ second}\).
Step 4: Formulating values for the graphs of Flux and Induced EMF. Let us model the equations mathematically over the intervals: • Interval \(0 \le t \le 1\text{ s}\) (Entry phase): The length of the loop inside the field increases as \(x(t) = v \cdot t\). \[ \Phi(t) = B \cdot A(t) = B \cdot l \cdot (vt) = B l v t \] The maximum flux occurs at \(t = 1\text{ s}\): \(\Phi_{\text{max}} = B l^2 = B \cdot (0.25)^2 = 0.0625 B\). The induced emf magnitude is: \[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (\text{constant value}) \] • Interval \(1\text{ s} < t \le 4\text{ s}\) (Fully immersed phase): The area inside the field remains constant and equal to the full area of the square, \(l^2\). \[ \Phi(t) = B \cdot l^2 = \Phi_{\text{max}} \quad (\text{constant flat line}) \] Since the flux is steady: \[ |\varepsilon| = \frac{d\Phi}{dt} = 0 \] • Interval \(4\text{ s} < t \le 5\text{ s}\) (Exit phase): The length inside decreases linearly. The remaining length inside is \(l - v(t - 4)\). \[ \Phi(t) = B \cdot l \cdot [l - v(t - 4)] \] This is a linearly decreasing line from \(\Phi_{\text{max}}\) down to \(0\). The induced emf magnitude during this exit window is: \[ |\varepsilon| = \left|\frac{d\Phi}{dt}\right| = Blv \quad (\text{constant value}) \]
Step 5: Graphical representation of the results. Based on our calculated functions, the plots look like this: • Magnetic Flux (\(\Phi\)) vs Time (\(t\)): • From \(t = 0\) to \(t = 1\text{ s}\): Straight line rising linearly from \(0\) to \(\Phi_{\text{max}}\). • From \(t = 1\text{ s}\) to \(t = 4\text{ s}\): Flat horizontal line staying at \(\Phi_{\text{max}}\). • From \(t = 4\text{ s}\) to \(t = 5\text{ s}\): Straight line falling linearly from \(\Phi_{\text{max}}\) to \(0\). • Magnitude of Induced EMF (\(|\varepsilon|\)) vs Time (\(t\)): • From \(t = 0\) to \(t = 1\text{ s}\): Constant positive horizontal step at a value of \(Blv\). • From \(t = 1\text{ s}\) to \(t = 4\text{ s}\): Drops down to zero and runs flat along the time axis. • From \(t = 4\text{ s}\) to \(t = 5\text{ s}\): Jumps back up to the constant positive horizontal step value of \(Blv\).
