Assignment 1, solutions

Problem 1:

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?


Motional emf
The conducting rod is moving in a plane perpendicular to B.
bulletDetails 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.

Problem 2:

(a)  Find the magnetic field at any point in the x-y plane for y > 0 due to a wire of length l carrying a current I from x = 0 to x = l along the x-axis.
(b)  Use the result of (a) to find the self inductance of a square current loop of side l.  You may leave your result as a definite integral, but the limits must be specified and the integrand must be in terms of the dimensions of the loop, constants, and the variables being integrated.
(c)  A conducting square current loop of side l with sides parallel to the x- and y-axis has resistance R and self inductance L.  It moves at a speed v in the +x-direction.  The loop passes through a magnetic field given by B = B0 k, -l/2< x < l/2, B = 0 elsewhere.  Find the current in the wire as a function of time assuming that I = 0 a long time before the loop reaches the magnetic field.


The  Biot-Savart law, self inductance, induced emf
We find the magnetic field due to the wire using the Biot-Savart law.  Given the magnetic field we use F = LI to find L.
bulletDetails of the calculation:
(a)  r = xi + yjB(r) = (μ0/(4π))∫ I dl'(r - r')/|r-r'|3.

dl'= dx'ir - r' = (x - x')i + yj|r - r'|3 = ((x - x')2 + y2)3/2.
dl'(r - r') = y dx' k.
B(r) = k0I/(4π))y∫0ldx' ((x - x')2 + y2)-3/2 =
k0I/(4π))y∫x-lxdx'' (x''2 + y2)-3/2 = k0I/(4πy))x''/(x''2 + y2)1/2|x-lx
= k0I/(4πy))[x/(x2 + y2)1/2 - (x - l)/((x - l)2 + y2)1/2].

(b)  F = LI,  F = ∫AB(r)∙n dA.

Flux due to side 1: F1 = ∫AB(r)∙k dA
= (μ0I/(4π))∫0l0ldx dy y-1[x/(x2 + y2)1/2 - (x - l)/((x - l)2 + y2)1/2].
L = (μ0I/π)∫0l0ldx dy y-1[x/(x2 + y2)1/2 - (x - l)/((x - l)2 + y2)1/2],
since the flux due to all sides if 4 times the flux due to 1 side.

(c)  ε = -dF/dt.  F = BA + LI.  (We let I be positive if it flows counterclockwise.)
ε = -B dA/dt - L dI/dt.
Let the right edge of the loop be at -l/2 at t = 0. 
For t < 0 the current I = 0.

Then for 0 < t < l/v we have
ε = -Blv - L dI/dt = IR,  dI/dt + IR/L + Blv/L = 0.  I = -Blv/R + C exp(-Rt/L).
I(0) = 0,  the constant C = Blv/R.  I(t) = -(Blv/R)(1 - exp(-Rt/L)).

For l/v < t < 2l/v we have
ε = Blv - L dI/dt = IR,  dI/dt + IR/L - Blv/L = 0.   I = Blv/R + C exp(-Rt/L).
I(l/v) = -(Blv/R)[1 - exp(-Rl/(Lv))] = Blv/R + C exp(-Rl/(Lv)).
C exp(-Rl/(Lv)) = -Blv/R[2 - exp(-Rl/(Lv))].
I(t') = Blv/R - Blv/R[2 - exp(-Rl/(Lv))]exp(-Rt'/L), with t' = t - l/v.

For t > 2l/v we have dI/dt + IR/L = 0.  I(t'') = I(2l/v) exp(-Rt''/L) with t'' = t - 2l/v.
I(2l/v) = Blv/R - Blv/R[2 - exp(-Rl/(Lv))]exp(-Rl/(Lv)).

Problem 3:

Consider the impedance bridge shown in the figure below.  Its purpose is to permit measurements of an unknown impedance Zu in terms of the fixed resistance RA and RB, variable resistance RS and variable capacitance CS.  If Zu is a pure resistance, then CS may be removed from the circuit (shorted out) and the impedance bridge becomes a simple Wheatstone bridge.
(a)  For a purely resistive impedance Zu find the value of RS for which a balance is obtained (no current on the null detector).
(b)  For a complex impedance Zu whose resistive component is Ru and whose reactive component is Xu, find the values of RS and CS for which balance is obtained.  Assume Zu has zero inductance. 
(c)  Show that for a purely inductive component Zu balance is not possible.


AC circuits
The circuit shown is a simple, two terminal, AC circuit.
bulletDetails of the calculation:
(a)  The circuit in part (a) is equivalent to the circuit shown below.

We want Vb = Va.
V = I1(RB + RS) = I2(RA + Ru),  We want I1RB = I2RA,  I1RS = I2Ru.
RS = (RB/RA)Ru.
(b)  The circuit in part (b) is equivalent to the circuit shown below.

We want Vb = Va.
We use the same arguments as in part (a) to show
ZS = (RB/RA)Zu = (RB/RA)(Ru + iXu)).
ZS = (RB/RA)Ru + iRBXu/RA) = RS - i/(ωCS).  RS = (RB/RA)Ru,  Cs = -RA/(RBXuω).
If Zu = (Ru + 1/(iωCu)) then Rs = (RB/RA)Ru,  Cs = RACu/RB.
(c)  If Zu = iωLu, we need RS - i/(ωCS) = iω(RB/RA)Lu, which is impossible, since all constants are positive.

Problem 4:

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.


Motional emf
The loop has no resistance.  The motional emf must be canceled by the induced emf due to the self inductance of the loop.
bulletDetails 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 5:

A DC motor has coils with a resistance of 30 Ω and operates from a voltage of 240 V.  When the motor is operating at its maximum speed, the peak back emf is 145 V.  
Find the peak current in the coils
(a)  when the motor is first turned on and
(b)  when the motor has reached maximum speed.
(c)  If the current in the motor is 6.0 A at some instant, what is the back emf at that time?


The motor has resistance and inductance.
bulletDetails of the calculation:
(a)  Immediately after the switch is closed, the motor coils are still stationary and the back emf is zero.
Thus I = ε/R = (240 V)/(30 Ω) = 8 A.
(b)  At maximum speed, εback = 145 V and
Ipeak =  (ε - εback)/R = (240 V - 145 V)/(30 Ω) =  3.2 A.
(c)  At a time when I = 6.0 A we have εback = ε - IR = 240 V - (6 A)(30 Ω) = 60 V.