We can consider an electron, in the moving wire. There is an electric field and a magnetic field acting on it in opposite directions, but they are the same in magnitude. So \( F_e = F_B \). We know the magnetic field is given by \( F = qvB \), and the electric force is \( F_e = Ee = \frac{Ve}{L} \). Equating these two formulas gives \( A \).
CloseFaraday's Law states that the magnitude of the induced emf is proportional to the rate of change of the magnetic flux through the conductor. This corresponds to option B.
CloseLenz's Law states that the direction of the induced current is such that it will oppose the change producing it. This means that the induced emf or current will always work to resist the change in flux, following Newton's third law of motion. This corresponds to option A.
CloseAccording to Lenz's law, when the magnet moves into the loop, a current is induced to oppose this change. Since the magnet moves to the right, increasing the magnetic field to the right in the loop, the loop will induce a current such that its magnetic field points to the left, to oppose the magnet's field. This means the loop will have its north pole on the left, next to the north pole of the magnet, which therefore will repel and move the loop to the right.
CloseAccording to Faraday's law, the induced emf is the rate of change of flux, which is exactly what a gradient describes. (This is quite trivial if you have learnt derivatives already)
CloseSince the speed increases, it will make 1 rotation in less time, hence the period decreases. Since the flux is changing faster too, the magnitude of the peak induced emf also increases, due to Faraday's law.
CloseMagnetic flux is given by: \[ \Phi = B A \cos\theta \] Substituting \( B = 0.5 \, \text{T} \), \( A = 0.02 \, \text{m}^2 \), and \( \cos 60^\circ = 0.5 \): \[ \Phi = (0.5)(0.02)(0.5) = 0.005 \, \text{Wb} \]
CloseFaraday's law states: \[ \mathcal{E} = -\frac{\Delta \Phi}{\Delta t} \] Substituting \( \Delta \Phi = 0.06 - 0.02 = 0.04 \, \mathrm{Wb} \) and \( \Delta t = 2 \, \mathrm{s} \): \[ \mathcal{E} = -\frac{0.04}{2} = -0.02 \, \mathrm{V} \] The magnitude of the emf is \( 0.02 \, \mathrm{V} \).
CloseThe induced emf is given by: \[ \mathcal{E} = B v L \] Substituting \( B = 0.8 \, \mathrm{T} \), \( v = 2.0 \, \mathrm{m/s} \), and \( L = 1.0 \, \mathrm{m} \): \[ \mathcal{E} = (0.8)(2.0)(1.0) = 1.6 \, \mathrm{V} \]
CloseThe maximum induced emf is given by: \[ \mathcal{E}_{\text{max}} = N B A \omega \] The angular velocity is: \[ \omega = 2\pi f = 2\pi (50) = 314 \, \text{rad/s} \] Substituting \( N = 100 \), \( B = 0.2 \, \text{T} \), \( A = 0.01 \, \text{m}^2 \), and \( \omega = 314 \): \[ \mathcal{E}_{\text{max}} = (100)(0.2)(0.01)(314) = 62.8 \, \text{V} \]
CloseThe correct answer is: C.
The maximum induced emf is proportional to the angular frequency: \[ \mathcal{E}_{\text{max}} = N B A \omega \] The angular frequency \( \omega \) is related to \( f \) by \( \omega = 2\pi f \). If \( f \) is doubled, \( \omega \) is also doubled, leading to a doubling of \( \mathcal{E}_{\text{max}} \).
CloseThe correct answer is: A.
According to Lenz's law, the induced current opposes the change in flux. Since the magnetic field into the page is decreasing, the induced current generates a field into the page to counteract this change. This corresponds to a clockwise current as viewed from above.
CloseThe correct answer is: B.
The frequency of the induced emf is equal to the rotational frequency of the coil. Since the coil rotates at \( 60 \: \mathrm{Hz} \), the frequency of the sinusoidal emf is also \( 60 \: \mathrm{Hz} \).
CloseThe correct answer is: D.
The induced emf is given by: \[ \mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} \] The change in magnetic flux \( \Delta \Phi \) is: \[ \Delta \Phi = A \Delta B = (0.01)(0.5 - 0) = 0.005 \: \mathrm{Wb} \] Substituting \( N = 300 \), \( \Delta \Phi = 0.005 \: \mathrm{Wb} \), and \( \Delta t = 2 \: \mathrm{ms} = 2 \times 10^{-3} \: \mathrm{s} \): \[ \mathcal{E} = -(300) \frac{0.005}{2 \times 10^{-3}} = 750 \: \mathrm{V} \]
Close| Option | Effect on Maximum emf |
|---|---|
| A | The emf remains the same. |
| B | The emf is halved. |
| C | The emf is doubled. |
| D | The emf becomes four times larger. |
| Option | Induced Current Direction |
|---|---|
| A | Clockwise |
| B | Counterclockwise |
| C | Alternating direction |
| D | No current is induced |
| Option | Frequency of Induced emf (Hz) |
|---|---|
| A | 30 |
| B | 60 |
| C | 120 |
| D | 240 |
| Option | Induced emf (V) |
|---|---|
| A | 500 |
| B | 250 |
| C | 100 |
| D | 750 |