University

Physics 2

Physics 2

Fall 2024

Consider the inductor shown. In each case, determine which end of the
inductor is at higher potential. State *right*, *left* or
*equal*.

- current flows to the left and is increasing
- current flows to the right and is decreasing
- current flows to the right with constant magnitude

- right
- right
- equal

Three \(LR\) circuits are made with the same resistor but different inductors. The figure shows the inductor current as a function of time.

Rank in order, from largest to smallest, the three inductances \(L_1, L_2, L_3\).

\(L_3 > L_2 > L_1\)

Time constant is \(\tau = L/R\). So greater \(L\) means longer time for current to decay.

For the circuit shown, the switch closes at time \(t=0\).

Describe each of the following as zero or at maximum value.

- battery current immediately after the switch closes
- battery current long after the switch closes
- inductor voltage immediately after the switch closes
- inductor voltage long after the switch closes
- resistor voltage immediately after the switch closes
- resistor voltage long after the switch closes

The inductor acts like an open circuit immediately after the switch closes, and acts like a short (zero resistance wire) long after.

- zero
- maximum
- maximum
- zero
- zero
- maximum

For each expression on the left, match the quantity on the right that has the same dimensions.

- \(\frac{1}{2} L I^2\)
- \(\displaystyle \frac{d\Phi_m}{dt}\)
- \(\displaystyle N \frac{\Phi_m}{I}\)
- \(\displaystyle \frac{L}{R}\)
- \(\displaystyle \frac{\mathcal{E}}{dI/dt}\)
- \(\displaystyle \frac{1}{\sqrt{LC}}\)
- \(\displaystyle \frac{\mu_0 N^2 A}{\ell}\)
- \(\displaystyle \frac{B^2}{2 \mu_0}\)

- voltage
- current
- inductance
- capacitance
- time
- frequency
- energy
- energy density

- energy
- voltage
- inductance
- time
- inductance
- frequency
- inductance
- energy density

These expressions may all be found on the equation sheet.

When the current in one coil changes at a rate of \(5.6\) A/s, an emf of \(6.3\) mV is induced in a second, nearby coil. What is the mutual inductance of the two coils?

\(\displaystyle M = \left | \frac{\mathcal{E}}{dI/dt} \right | = 1.125\) mH

(Review section 14.1 Mutual Inductance).

For the circuit shown below, \(\mathcal{E} = 50\) V, \(R_1=10\) Ω, \(R_2=R_3=19.4\) Ω, and \(L=2.0\) mH.

Switch \(S\) has been open for a long time.

At time \(t=0\) switch \(S\) is closed.

At time \(t=t_1\), a long time later, \(S\) is opened.

Find the values of \(I_1\), \(I_2\) and \(I_3\) at the following times. (Report negative current when it flows opposite the direction indicated in the figure.)

- \(t=0^+\) (immediately after \(t=0\))
- \(t=t_1^-\) (immediately before \(t=t_1\))
- \(t=t_1^+\)
- \(t=t_1 + 20\) μs
- a long time after \(t=t_1\)

- \(I_1= I_2 = 1.70\) A; \(I_3=0\)
- \(I_1 = 2.54\) A; \(I_2 = 1.27\) A; \(I_3 = 1.27\) A
- \(I_1 = 0\); \(I_2 = -1.27\) A; \(I_3 = 1.27\) A
- \(I_1 = 0\); \(I_2 = -0.861\) A; \(I_3 = 0.861\) A
- \(I_1= I_2 = I_3 = 0\)

Reasoning:

The equivalent circuits for parts a-d are as follows:

c-d.

- For this we need exponential decay with time constant \(\tau = \frac{L}{R}\), where \(R = R_2 + R_3\). Also, since the arrow for \(I_2\) points down, we require \(I_2 = - I_3\).

(Review section 14.4 RL Circuits).

An \(LC\) circuit in an AM tuner (in a car stereo) uses a coil with an inductance of \(2.5\) mH and a variable capacitor. The natural frequency of the circuit is to be adjustable over the range \(540\) to \(1600\) kHz (the AM broadcast band).

What range of capacitance is required?

\(C\) must be in the range \(34.7\) pF to \(3.96\) pF.

Since \(\displaystyle \omega = \sqrt{ \frac{1}{LC}}\) and \(\displaystyle \omega = 2 \pi f\), we need \(\displaystyle C = \frac{1}{4 \pi^2 f^2 L}\).

(Review section 14.5 Oscillations in an LC Circuit).

In the circuit shown below, \(S_1\) is opened and \(S_2\) is closed simultaneously.

Determine the following:

- the frequency of the resulting oscillations,
- the maximum charge on the capacitor,
- the maximum current through the inductor, and
- the electromagnetic energy of the oscillating circuit.

- 252 Hz
- \(48\) μC
- \(76\) mA
- \(2.9 \times 10^{-4}\) J

Reasoning:

- \(\displaystyle f = \frac{\omega}{2 \pi} = \frac{2 \pi}{\sqrt{LC}}\)
- \(\displaystyle Q = CV_{bat}\)
- All quantities in an \(LC\) circuit oscillate sinusoidally with frequency \(\omega\). We know the peak charge is \(Q_0 = 48\) μC, so the charge at all times must be \(\displaystyle q(t) = Q_0 \cos (\omega t + \phi)\). Current is the derivative of this: \(\displaystyle i(t) = dq/dt = -Q_0 \omega \sin (\omega t + \phi)\), so the maximum current is \(I_0 = Q_0 \omega\).
- Energy can be found from either \(U = \frac{1}{2} C V_0^2\) or \(U = \frac{1}{2} L I_0^2\).

(Review section 14.5 Oscillations in an LC Circuit).

*Last modified: Tue April 16 2024, 06:30 PM.*