A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the center of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π² Ω, then the total charge flowing through the coil during this time is:
32 μC
Understanding the Physics Problem
This problem involves a long solenoid and a smaller coil placed inside it. The current in the solenoid changes over time, which causes the magnetic field produced by the solenoid to change. This changing magnetic field links with the coil, inducing an electromotive force (emf) and hence a current in the coil. We need to find the total electric charge that flows through the coil during the time the solenoid's current is changing.
The key concepts involved are the magnetic field of a solenoid, magnetic flux, Faraday's Law of electromagnetic induction, Ohm's law, and the definition of electric charge flow.
Key Concepts for Electromagnetic Induction
- Magnetic Field of a Long Solenoid: The magnetic field inside a long solenoid is approximately uniform and parallel to the axis, given by \( B = \mu_0 n I \), where \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current in the solenoid.
- Magnetic Flux: The magnetic flux (\(\Phi\)) through a loop is the measure of the total magnetic field lines passing through the loop's area. If the field is uniform and perpendicular to the area, \(\Phi = B A\). For a coil with \(N\) turns, the total flux is \( \Phi_{total} = N B A \).
- Faraday's Law of Induction: The induced emf (\(\mathcal{E}\)) in a coil is proportional to the rate of change of magnetic flux through it: \( \mathcal{E} = -\frac{d\Phi_{total}}{dt} \). The negative sign indicates the direction of the induced current (Lenz's Law).
- Induced Current and Charge: If the coil has resistance \(R\), the induced current is \( i = \frac{\mathcal{E}}{R} \). The total charge \(Q\) flowing through the coil during a time interval \(\Delta t\) is the integral of the current over that time: \( Q = \int i \, dt \). Using Faraday's Law, \( Q = \int \frac{|\mathcal{E}|}{R} dt = \int \frac{1}{R} \left| -\frac{d\Phi_{total}}{dt} \right| dt \). If the resistance is constant, this simplifies to \( Q = \frac{1}{R} \int |d\Phi_{total}| = \frac{|\Delta \Phi_{total}|}{R} \), where \(|\Delta \Phi_{total}|\) is the magnitude of the change in magnetic flux.
Step-by-Step Calculation of Total Charge
Let's identify the given parameters:
- Solenoid turns per meter, \(n = 2 \times 10^4\) m\(^{-1}\).
- Solenoid diameter is 0.1 m, but this information is not needed for the magnetic field inside a long solenoid.
- Coil turns, \(N_{coil} = 100\).
- Coil radius, \(r_{coil} = 0.01\) m.
- Coil area, \(A_{coil} = \pi r_{coil}^2 = \pi (0.01)^2 = \pi \times 10^{-4}\) m\(^2\).
- Initial current in solenoid, \(I_{initial} = 4\) A.
- Final current in solenoid, \(I_{final} = 0\) A.
- Time interval, \(\Delta t = 0.05\) s.
- Resistance of the coil, \(R_{coil} = 10\pi^2 \, \Omega\).
- Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T m/A.
We will use the relationship \( Q = \frac{|\Delta \Phi_{total}|}{R_{coil}} \) to find the total charge. First, let's calculate the total flux through the coil at the initial and final times.
The magnetic field inside the solenoid is given by \( B = \mu_0 n I \). Since the coil is placed at the center and its axis coincides with the solenoid axis, the magnetic field is perpendicular to the area of each turn of the coil. The area of the coil is \( A_{coil} = \pi r_{coil}^2 \).
The total magnetic flux through the coil with \(N_{coil}\) turns is \( \Phi_{total} = N_{coil} B A_{coil} = N_{coil} (\mu_0 n I) (\pi r_{coil}^2) \).
Initial flux through the coil (\(I = I_{initial}\)):
\[ \Phi_{initial} = N_{coil} \mu_0 n I_{initial} \pi r_{coil}^2 \] \[ \Phi_{initial} = 100 \times (4\pi \times 10^{-7} \text{ T m/A}) \times (2 \times 10^4 \text{ m}^{-1}) \times (4 \text{ A}) \times \pi (0.01 \text{ m})^2 \] \[ \Phi_{initial} = 100 \times 4\pi \times 10^{-7} \times 2 \times 10^4 \times 4 \times \pi \times 10^{-4} \] \[ \Phi_{initial} = (100 \times 4 \times 2 \times 4) \times (\pi \times \pi) \times (10^{-7} \times 10^4 \times 10^{-4}) \] \[ \Phi_{initial} = 3200 \pi^2 \times 10^{-7} \text{ Wb} \]Final flux through the coil (\(I = I_{final}\)):
\[ \Phi_{final} = N_{coil} \mu_0 n I_{final} \pi r_{coil}^2 \] Since \(I_{final} = 0\) A, the final magnetic field is 0, and thus the final flux is 0. \[ \Phi_{final} = 0 \text{ Wb} \]The change in magnetic flux is \(\Delta \Phi_{total} = \Phi_{final} - \Phi_{initial}\):
\[ \Delta \Phi_{total} = 0 - 3200 \pi^2 \times 10^{-7} \text{ Wb} = -3200 \pi^2 \times 10^{-7} \text{ Wb} \]The magnitude of the change in flux is \(|\Delta \Phi_{total}| = 3200 \pi^2 \times 10^{-7}\) Wb.
Now, calculate the total charge flowing through the coil using the resistance \(R_{coil} = 10\pi^2 \, \Omega\):
\[ Q = \frac{|\Delta \Phi_{total}|}{R_{coil}} \] \[ Q = \frac{3200 \pi^2 \times 10^{-7} \text{ Wb}}{10\pi^2 \, \Omega} \] \[ Q = \frac{3200 \pi^2}{10\pi^2} \times 10^{-7} \text{ C} \] \[ Q = 320 \times 10^{-7} \text{ C} \] \[ Q = 32 \times 10^{-6} \text{ C} \] \[ Q = 32 \, \mu\text{C} \]Thus, the total charge flowing through the coil during this time is \(32 \, \mu\text{C}\).
Result
The calculated total charge flowing through the coil is \(32 \, \mu\text{C}\).
Revision Table: Electromagnetic Induction Calculations
| Parameter | Symbol | Formula/Value | Unit |
|---|---|---|---|
| Solenoid turns per meter | \(n\) | \(2 \times 10^4\) | m\(^{-1}\) |
| Coil turns | \(N_{coil}\) | 100 | |
| Coil radius | \(r_{coil}\) | 0.01 | m |
| Coil Area | \(A_{coil}\) | \(\pi r_{coil}^2 = \pi \times 10^{-4}\) | m\(^2\) |
| Initial Solenoid Current | \(I_{initial}\) | 4 | A |
| Final Solenoid Current | \(I_{final}\) | 0 | A |
| Permeability of Free Space | \(\mu_0\) | \(4\pi \times 10^{-7}\) | T m/A |
| Magnetic Field in Solenoid | \(B\) | \(\mu_0 n I\) | T |
| Total Magnetic Flux through Coil | \(\Phi_{total}\) | \(N_{coil} B A_{coil}\) | Wb |
| Change in Flux Magnitude | \(|\Delta \Phi_{total}|\) | \(|N_{coil} \mu_0 n \pi r_{coil}^2 (I_{final} - I_{initial})|\) | Wb |
| Coil Resistance | \(R_{coil}\) | \(10\pi^2\) | \(\Omega\) |
| Total Charge | \(Q\) | \(\frac{|\Delta \Phi_{total}|}{R_{coil}}\) | C |
Additional Information: Induced Current and Charge Flow
The induced current \(i(t)\) in the coil is given by \( i(t) = \frac{\mathcal{E}(t)}{R} = -\frac{1}{R} \frac{d\Phi_{total}}{dt} \). The total charge \(Q\) that flows during the time interval from \(t_1\) to \(t_2\) is:
\[ Q = \int_{t_1}^{t_2} i(t) \, dt = \int_{t_1}^{t_2} -\frac{1}{R} \frac{d\Phi_{total}}{dt} \, dt \]Since the resistance \(R\) is constant, we can write:
\[ Q = -\frac{1}{R} \int_{\Phi_{total}(t_1)}^{\Phi_{total}(t_2)} d\Phi_{total} = -\frac{1}{R} [\Phi_{total}(t)]_{t_1}^{t_2} = -\frac{1}{R} (\Phi_{total}(t_2) - \Phi_{total}(t_1)) = -\frac{\Delta \Phi_{total}}{R} \]The total charge flow is related to the total change in magnetic flux. The sign of the charge simply indicates the direction of flow based on the direction of flux change. When asked for the total charge flowing, the magnitude is usually implied, especially in multiple-choice options where direction is not specified.
In this problem, the current decreases linearly with time, meaning \( \frac{dI}{dt} \) is constant. Consequently, \( \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \) is also constant. Since \( \Phi_{total} = N_{coil} A_{coil} B \), \( \frac{d\Phi_{total}}{dt} = N_{coil} A_{coil} \frac{dB}{dt} \) is constant. This means the induced emf \( \mathcal{E} \) and induced current \( i \) are constant in magnitude during the time interval \(\Delta t\). In this case, the total charge could also be calculated as \( Q = |i| \Delta t \), where \(|i| = \frac{|\mathcal{E}|}{R} = \frac{|-N_{coil} A_{coil} \mu_0 n (dI/dt)|}{R}\). The rate of change of current is \( \frac{dI}{dt} = \frac{0 \, A - 4 \, A}{0.05 \, s} = \frac{-4}{0.05} \, A/s = -80 \, A/s \). The magnitude of induced emf is \( |\mathcal{E}| = |100 \times \pi \times 10^{-4} \times 4\pi \times 10^{-7} \times 2 \times 10^4 \times (-80)| \) \( |\mathcal{E}| = |100 \times \pi \times 10^{-4} \times 8\pi \times 10^{-3} \times (-80)| \) \( |\mathcal{E}| = |800\pi^2 \times 10^{-7} \times (-80)| \) \( |\mathcal{E}| = 64000 \pi^2 \times 10^{-7} = 6.4 \pi^2 \times 10^{-3}\) V. The magnitude of induced current is \( |i| = \frac{6.4 \pi^2 \times 10^{-3} \, V}{10 \pi^2 \, \Omega} = \frac{6.4}{10} \times 10^{-3} \, A = 0.64 \times 10^{-3} \, A \). The total charge is \( Q = |i| \Delta t = (0.64 \times 10^{-3} \, A) \times (0.05 \, s) = 0.032 \times 10^{-3} \, C = 32 \times 10^{-6} \, C = 32 \, \mu C \). Both methods yield the same result, confirming the calculation.
Important Questions from Electromagnetic Induction
In a pair of adjacent coils, for a change of current in one of the coils from 0 A to 10 A in 0.25 s, the magnetic flux in the adjacent coil changes by 15 Wb. The mutual inductance of the coils is:
A 50 Hz AC current of crest value 1 A flows through the primary of a transformer. If the mutual inductance between the primary and secondary is 0.5 H, the crest voltage induced in the secondary is:
Lower half of a convex lens is made opaque. Which of the following statements describes the image of the object placed in front of the lens?
A transformer has an efficiency of 80%. It works at 3 kW and 120 V. If the secondary voltage is 240 V, what will be the secondary current?
In an AC generator when the plane of the armature is perpendicular to the magnetic field, what will the magnitude of the magnetic flux passing through the coil and the emf induced in the coil be?
