Induction - AP Physics 2

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Question

A magnetic rod is inside a coil of wire which is connected to an ammeter. If the rod is stationary, which of the following statements are true?

Answer

Current is induced in wire when the magnetic flux changes. When the magnetic rod is in motion, the flux is changing, so current is induced. If the coil were expanded or contracted with the rod still there, the flux would change and current would be induced. In our case, the rod is stationary and the coil isn't changing shape. Therefore, the flux is not changing, so there is no current being induced. Additionally, there's no reason for the rod to lose its magnetic property.

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Question

Suppose the induced electromotive force of a double loop wire has a magnitude of $1.1\textup{V}$ when the change in magnetic flux is $0.683\textup{T} \textup{m}^2$ . How much time has elapsed for the flux change?

Answer

Write the formula for induced electromotive force.

$\left |\epsilon \right |=N\left | \frac{\Phi_{final}-\Phi_{initial} }{t_{final}-t_{initial} } \right |=N\left | \frac{\Delta \Phi }{\Delta t} \right |$

Since there are two loops, .

$\left |\epsilon \right |= 1.1\textup{V}$

$\Delta \Phi=0.683 \textup{T} \textup{m}^2$

Solve for $\Delta t$ .

$\Delta t=N\left | \frac{\Delta \Phi}{\epsilon } \right | = 2\left | \frac{0.683 \textup{T} \textup{m}^2}{1.1\textup{V}} \right |=1.24 \textup{s}$

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Question

A system consists of two rails, whose resistance is zero, a conductive bar, whose resistance is also zero, and a resistor whose resistance, 0.02 Ohms. The bar remains in contact with the rails as it slides to the right with a velocity $v=22\frac{m}{s}$ . The distance from one rail to the other is 0.07m. The magnetic field, which is directed out of the page, has a magnitude of 0.0015T. What is the magnitude and direction of current in the resistor?

Answer

When a conductor moves through a magnetic field in such a way that it cuts through magnetic field lines, the mobile charge carriers separate due to the magnetic force on them, creating a potential . Since there is no resistance anywhere else in the circuit, all of this potential is lost in the resistor, so we can apply Ohm's law:

$I=\frac{V}{R}$

$I=\frac{Blv}{R}$

$I=\frac{0.0015T0.07m22\frac{m}{s}}{0.02\Omega }=0.1155A$

Because the positive charge in the rod feels an upward force due to the right-hand rule, the top of the rod has a greater potential than the bottom, and current flows counterclockwise around the circuit, resulting in a downward direction in the resistor.

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Question

A $150\Omega$ resistor is connected to a coil with turns and a cross sectional area of $0.02m^{2}$ . A magnet is lowered as shown in the figure. The magnetic field increases from to in . Find the magnitude of the current going through the resistor.

Answer

There is a change in flux produced by the changing magnetic field which is given by

$\Delta \Phi=A \Delta B \cos\left ( \theta\right )$

where is the cross sectional area of the coil, $\Delta B$ is the change in magnetic field, and $\theta$ is the angle of the field lines relative to the normal of the cross section of the coil.

In this case the magnet is perpendicular to the cross section of the coil and so

$\theta=0 \rightarrow \cos\left ( 0\right )=1$

$\Delta \Phi=A \Delta B$

The change in magnetic field is just the final given value minus the intial value. Faraday's Law says that an emf will be generated by a change in flux,

$\varepsilon=N\left \left | \frac{\Delta \Phi}{\Delta t}\right |$

where is the number of turns in the coil. Plugging in the change in flux gives

$\varepsilon=N \left | A\frac{\Delta B}{\Delta t}\right |$

The change in time is just since we can start our clock at zero. The current can be found using Ohm's Law where

$I=\frac{\varepsilon}{R}=\frac{N A \Delta B}{R \Delta t} =0.01A$

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