Consider the circuit shown. At t = 0 the switch is closed.

(a) What are the values of the currents through the resistors R_{1}, R_{2},
and R_{3} just after the switch has been closed and a long time after
the switch has been closed?

(b) Find the currents through the resistors R_{1}, R_{2}, and R_{3}
as a function of time for t > 0.

Solution:

- Concepts:

Kirchhoff's rules, resistors in series and parallel - Reasoning:

We have a circuit with transients. As t --> ∞, the currents will become constant. - Details of the calculation:

(a) Let us denote the constant currents as t --> ∞ with the subscript f.

I_{1f}= V/R_{eff}, R_{eff}= R_{1}+ R_{2}R_{3}/(R_{2}+ R_{3}) = (R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3})/(R_{2}+ R_{3})

I_{2f}= V(1 - R_{1}/R_{eff})/R_{2}.

I_{3f}= V(1 - R_{1}/R_{eff})/R_{3}.

Just after the switch has been closed I_{10}= I_{20}= V/(R_{1}+ R_{2}), I_{30}= 0.

(b) Kirchhoff's rules: We write down 3 equations for 3 unknowns.

(1) V - R_{1}I_{1}- R_{2}I_{2}= 0,

(2) -I_{2}R_{2}+ I_{3}R_{3}+ LdI_{3}/dt = 0,

(3) I_{1}- I_{2}- I_{3}= 0.

Try solutions of the form

I_{2}= I_{2f}+ (I_{20}- I_{2f})exp(-At), I_{3}= I_{3f}(1 - exp(-At)), dI_{3}/dt = AI_{3f}exp(-At).

Check that they satisfy equations 1 - 3, and solve for A.

equ 3: I_{1}= I_{2f}+ (I_{20}- I_{2f})exp(-At) + I_{3f}(1 - exp(-At)).

At t = 0 this yields the correct values for I_{1}.

equ 2: -R_{2}(I_{2f}+ (I_{20}- I_{2f})exp(-At)) + R_{3}(I_{3f}(1 - exp(-At))) + LAI_{3f}exp(-At) = 0

for all t > 0.

This implies that -R_{2}I_{20}+ R_{2}I_{2f}- R_{3}I_{3f}+ LAI_{3f}= 0,

A = (R_{2}I_{20})/( LI_{3f}) = (R_{eff}/L)(R_{2}+ R_{3})/(R_{1}+ R_{2}).

equ 1: R_{1}I_{1}+ R_{2}I_{2}= V.

This implies that

R_{1}(I_{20}- I_{2f}) + R_{1}I_{3f}+ R_{2}(I_{20}- I_{2f}) = 0, or R_{1}I_{2f}- R_{1}I_{3f}+ R_{2}I_{2f}= V, which holds.

[R_{1}I_{2f}- R_{1}I_{3f}+ R_{2}I_{2f}= R_{1}V(_{1}/R_{1}+ 1/R_{2}+ 1/R_{3})(1 - R_{1}/R_{eff})

= (V/R_{eff})(( R_{1}R_{2}+ R_{1}R_{3}+ R_{2}R_{3})/(R_{2}R_{3}))( R_{2}R_{3}/(R_{2}+ R_{3})) = V]

The trial solutions satisfy Kirchhoff's rule for t > 0 with a time constant

τ = 1/A = (L/R_{eff})(R_{1}+ R_{2})/(R_{2}+ R_{3}).

A metal bar of mass m slides without friction on two parallel conducting
rails a distance l apart (see figure). A resistor R is connected across
the rails and a uniform magnetic field **B**, pointing into the page, fills
the entire region.

(a) If the bar moves to the right at speed v, what is the current in the
resistor? Indicate also in what direction the current flows.

(b) What is the magnetic force on the bar? Indicate the direction of the
force and provide the result in terms of the data given.

(c) If the bar starts out with speed v_{0} at time t = 0, and is left to
slide, what is its speed at a later time t?

(d) The initial kinetic energy of the bar was, of course, mv_{0}^{2}/2.
Where does this energy go? Prove that energy is conserved in this process
by showing that the energy gained elsewhere is exactly mv_{0}^{2}/2.

Solution:

- Concepts:

Motional emf - Reasoning:

The emf is generated by sliding the conducting rod in a direction perpendicular to the magnetic field - Details of the calculation:

(a) The emf driving the current is equal to vBl, where l is the length of the rod. (The work done per unit charge (voltage) is vBl, when a charge moves from one end of the moving wire to the other end.) The current through the resistor is I = vBl/R, it flows counterclockwise.

(b) The force on the bar has magnitude F = IlB = B^{2}l^{2}v/R and points towards the left.

(c) F = -B^{2}l^{2}v/R = mdv/dt. dv/dt = -B^{2}l^{2}v/(mR), v(t) = v_{0}exp(-t/τ), τ = mR/(B^{2}l^{2}).

(c) Kinetic energy T = ½Mv^{2}= ½mv_{0}^{2}exp(-2t/τ),

dT/dt = -(mv_{0}^{2}/τ)exp(-2t/τ) = -(v_{0}^{2}B^{2}l^{2}/R)exp(-2t/τ).

The rate of ohmic heating per unit volume is P = I^{2}R = v_{0}^{2}B^{2}l^{2}exp(-2t/τ)/R.

dT/dt = -P. Kinetic energy is converted into thermal energy.

As t --> ∞, the total amount of thermal energy produced is

∫_{0}^{∞}Pdt = (mv_{0}^{2}/τ) ∫_{0}^{∞}exp(-2t/τ)dt = ½mv_{0}^{2}.

A^{ }square loop made of wire with negligible resistance is placed^{
}on a horizontal frictionless table as shown (top view). The^{ }
mass of the loop is m and the length of^{ }each side is b. A
non-uniform vertical magnetic field exists^{ }in the region; its
magnitude is given by the formula^{ }B = B_{0} (1 + kx), where B_{0}
and k are known constants.^{ }

The loop is^{ }given a quick push with an initial velocity v along^{
}the x-axis as shown. The loop stops after a time^{ }interval t.
Find the self-inductance L of the loop.^{ }

Solution:

- Concepts:

Motional emf - Reasoning:

The loop has no resistance. The motional emf must be canceled by the induced emf due to the self inductance of the loop. - Details of the calculation:

Let v(t) be positive, if it points into the positive x-direction and let I(t) be positive if it flows clockwise in the loop. When the loop moves with positive velocity v(t), the motional emf = v(t)bB(x '+ b) - v(t)bB(x') = v(t)b^{2}B_{0}k causes a current I(t) to flow clockwise. Since the loop has no resistance, Kirchhoff's loop rule requires vb^{2}B_{0}k - LdI/dt = 0. The motional emf is canceled by the induced emf due to the self inductance of the loop. We have

(1) vb^{2}B_{0}k = LdI/dt.

The magnetic force on the current carrying loop is

F = mdv/dt = -IbB(x' + b) + IbB(x') = -Ib^{2}B_{0}k.

The acceleration points in the negative x-direction if the direction of current flow is clockwise. We have

(2) dv/dt = -Ib^{2}B_{0}k/m

We can differentiate (1) with respect to t and obtain

(3) dv/dt = (L/(b^{2}B_{0}k))d^{2}I/dt^{2}.

Combining (2) and (3) we obtain

(4) d^{2}I/dt^{2}= -[(b^{2}B_{0}k)^{2}/(mL)] I.

The solution to this second-order differential equation satisfying I(t = 0) = 0 is

I(t) = I_{max}sin(ωt), with ω^{2}= (b^{2}B_{0}k)^{2}/(mL).

The velocity v is proportional to dI/dt, so v(t) = v_{max}cos(ωt).

The loop oscillates about its starting position along the x-direction, v(t) = 0 for the first time for ωt = π/2. This yields L = (1/m)(2b^{2}B_{0}kt/π)^{2}in terms of the given parameters.

In order to suppress the 120 Hz hum from the power supply rectifier or
amplifier, a "smoothing filter" is used. In its simplest form it
consists of a resistance in series with a capacitance, as shown in the figure.
If the applied voltage has a DC component V_{0} and a 120 cycle
component of amplitude V_{2}, find the corresponding voltages at the
terminals of the capacitor for R = 10^{3 }Ω and C = 10 μF.

Solution:

- Concepts:

AC circuits - Reasoning:

We treat the circuit as a single loop AC circuit containing a capacitor. We are asked to find the voltage across one of the circuit elements. The DC voltage component is then added to the AC component. - Details of the calculation:

Z(capacitance) = Z_{C}= 1/(iωC), Z(resistance) = Z_{R }= R.

I = V_{AC_in}/(Z_{C}+ Z_{R}), V_{AC_out}= IZ_{C}= V_{AC_in}Z_{C}/(Z_{C}+ Z_{R}).

Z_{C}/(Z_{C}+ Z_{R}) = 1/(1 + iRωC) = (1 - iRωC)/(1 + ω^{2}R^{2}C^{2}) = [1/(1 + ω^{2}R^{2}C^{2})^{1/2}]e^{-iφ}.

tanφ = RωC.

V_{out}= V_{0 }+ V_{AC_out}= V_{0}+ V_{2}[1/(1 + ω^{2}R^{2}C^{2})^{1/2}]e^{i(ωt-φ)}.

Here ω = 2π*120 s^{-1}, R = 10^{3}Ω, C = 10^{-5}F.

V_{out}= V_{0 }+ 0.13*V_{2}e^{i(ωt-φ)}.

Refer to the figure. One end of a conducting rod rotates with angular
velocity ω in a circle of radius a making contact with a horizontal, conducting
ring of the same radius. The other end of the rod is fixed. Stationary
conducting wires connect the fixed end of the rod (A) and a fixed point on the
ring (C) to either end of a resistance R. A uniform vertical magnetic field
**B** passes through the ring.

(a) Find the current I flowing through the resistor and the rate at which
heat is generated in the resistor.

(b) What is the sign of the current, if positive I corresponds to flow in the
direction of the arrow in the figure?

(c) What torque must be applied to the rod to maintain its rotation at the
constant angular rate ω?

What is the rate at which mechanical work must be done?

Solution:

- Concepts:

Motional emf - Reasoning:

The conducting rod is moving in a plane perpendicular to**B**. - Details of the calculation:

(a) The conducting rod is moving in a plane perpendicular to**B**.

Speed of the rod as a function of the distance r from the origin: v(r) = ωr.

Force on an electron: F_{e}= q_{e}vB = q_{e}Bωr.

Work done per unit charge: emf = Bω∫_{0}^{a}rdr = Bωa^{2}/2.

Assume that the resistance of the conducting rod and the wires is negligible.

I = emf/R = Bωa^{2}/(2R) is the current flowing through the resistor.

P_{e}= I^{2}R = B^{2}ω^{2}a^{4}/(4R) is the rate heat is generated.

(b) The sign of the current is positive.

(c) Force on a section dr of the current carrying rod: dF = IdrB (direction clockwise in the figure).

An external force of equal magnitude and opposite direction is needed to maintain the constant angular speed.

dτ = rdF = r IBdr, τ = IB∫_{0}^{a}rdr = IBa^{2}/2 = B^{2}ωa^{4}/(4R).

The rate at which mechanical work is done is P_{m}= τω = B^{2}ω^{2}a^{4}/(4R) = P_{e}.