## Assignment 1, solutions

#### Problem 1:

Consider the circuit shown.  At t = 0 the switch is closed.
(a)  What are the values of the currents through the resistors R1, R2, and R3 just after the switch has been closed and a long time after the switch has been closed?
(b)  Find the currents through the resistors R1, R2, and R3 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.
I1f = V/Reff,  Reff = R1 + R2R3/(R2 + R3) = (R1R2 + R1R3 + R2R3)/(R2 + R3)
I2f = V(1 - R1/Reff)/R2.
I3f = V(1 - R1/Reff)/R3.
Just after the switch has been closed I10 = I20 = V/(R1 + R2),  I30 = 0.
(b)  Kirchhoff's rules:  We write down 3 equations for 3 unknowns.
(1)  V - R1I1 - R2I2 = 0,
(2)  -I2R2 + I3R3 + LdI3/dt = 0,
(3)  I1 - I2 - I3 = 0.
Try solutions of the form
I2 = I2f + (I20 - I2f)exp(-At),  I3 = I3f(1 - exp(-At)),  dI3/dt = AI3fexp(-At).
Check that they satisfy equations 1 - 3, and solve for A.
equ 3:  I1 = I2f + (I20 - I2f)exp(-At) + I3f(1 - exp(-At)).
At t = 0 this yields the correct values for I1.
equ 2:  -R2(I2f + (I20 - I2f)exp(-At)) + R3(I3f(1 - exp(-At))) + LAI3fexp(-At) = 0
for all t > 0.
This implies that -R2I20 + R2I2f - R3I3f + LAI3f = 0,
A = (R2I20)/( LI3f) = (Reff/L)(R2 + R3)/(R1 + R2).
equ 1:  R1I1 + R2I2 = V.
This implies that
R1(I20 - I2f) + R1I3f + R2(I20 - I2f) = 0, or R1I2f - R1I3f + R2I2f = V, which holds.
[R1I2f - R1I3f + R2I2f = R1V(1/R1 + 1/R2 + 1/R3)(1 - R1/Reff)
= (V/Reff)(( R1R2 + R1R3 + R2R3)/(R2R3))( R2R3/(R2 + R3)) = V]
The trial solutions satisfy Kirchhoff's rule for t > 0 with a time constant
τ = 1/A = (L/Reff)(R1 + R2)/(R2 + R3).

#### Problem 2:

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 v0 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, mv02/2.  Where does this energy go?  Prove that energy is conserved in this process by showing that the energy gained elsewhere is exactly mv02/2.

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 = B2l2v/R and points towards the left.
(c)  F = -B2l2v/R = mdv/dt.  dv/dt = -B2l2v/(mR),  v(t) = v0exp(-t/τ),  τ = mR/(B2l2).
(c)  Kinetic energy T = ½Mv2 = ½mv02exp(-2t/τ),
dT/dt = -(mv02/τ)exp(-2t/τ) = -(v02B2l2/R)exp(-2t/τ).
The rate of ohmic heating per unit volume is P = I2R = v02B2l2exp(-2t/τ)/R.
dT/dt = -P.  Kinetic energy is converted into thermal energy.
As t --> ∞, the total amount of thermal energy produced is
0Pdt = (mv02/τ) ∫0exp(-2t/τ)dt = ½mv02.

#### Problem 3:

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 = B0 (1 + kx), where B0 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)b2B0k causes a current I(t) to flow clockwise.  Since the loop has no resistance, Kirchhoff's loop rule requires vb2B0k - LdI/dt = 0.  The motional emf is canceled by the induced emf due to the self inductance of the loop.  We have
(1)  vb2B0k = LdI/dt.
The magnetic force on the current carrying loop is
F = mdv/dt = -IbB(x' + b) + IbB(x') = -Ib2B0k.
The acceleration points in the negative x-direction if the direction of current flow is clockwise.  We have
(2)  dv/dt = -Ib2B0k/m
We can differentiate (1) with respect to t and obtain
(3)  dv/dt = (L/(b2B0k))d2I/dt2.
Combining (2) and (3) we obtain
(4)  d2I/dt2 = -[(b2B0k)2/(mL)] I.
The solution to this second-order differential equation satisfying I(t = 0) = 0 is
I(t) = Imaxsin(ωt), with ω2 = (b2B0k)2/(mL).
The velocity v is proportional to dI/dt, so v(t) = vmaxcos(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)(2b2B0kt/π)2 in terms of the given parameters.

#### Problem 4:

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 V0 and a 120 cycle component of amplitude V2, find the corresponding voltages at the terminals of the capacitor for R = 103 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) = ZC = 1/(iωC),  Z(resistance) = ZR = R.
I = VAC_in/(ZC + ZR),  VAC_out = IZC = VAC_inZC/(ZC + ZR).
ZC/(ZC + ZR) = 1/(1 + iRωC) = (1 - iRωC)/(1 + ω2R2C2) = [1/(1 + ω2R2C2)1/2]e-iφ.
tanφ   = RωC.
Vout = V0 + VAC_out = V0 + V2[1/(1 + ω2R2C2)1/2]ei(ωt-φ).
Here ω = 2π*120 s-1, R = 103 Ω, C = 10-5 F.
Vout = V0 + 0.13*V2ei(ωt-φ).

#### Problem 5:

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: Fe = qevB = qeBωr.
Work done per unit charge: emf = Bω∫0ardr = Bωa2/2.
Assume that the resistance of the conducting rod and the wires is negligible.
I = emf/R = Bωa2/(2R) is the current flowing through the resistor.
Pe = I2R = B2ω2a4/(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∫0ardr = IBa2/2 = B2ωa4/(4R).
The rate at which mechanical work is done is Pm = τω =  B2ω2a4/(4R) = Pe.