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}. |

(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** = B_{0} **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.

Solution:

Concepts: The Biot-Savart law, self inductance, induced emf | |

Reasoning: 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. | |

Details of the calculation: (a) r = xi + yj. B(r) = (μ_{0}/(4π))∫_{
}I dl'×(r - r')/|r-r'|^{3}.d l'= dx'i, r - r' = (x - x')i + yj,
|r -
r'|^{3} = ((x - x')^{2} + y^{2})^{3/2}.d l'×(r - r') = y dx' k.B(r) = k (μ_{0}I/(4π))y∫_{0}^{l}dx'
((x - x')^{2} + y^{2})^{-3/2} =k (μ_{0}I/(4π))y∫_{x-l}^{x}dx'' (x''^{2}
+ y^{2})^{-3/2} = k (μ_{0}I/(4πy))x''/(x''^{2}
+ y^{2})^{1/2}|_{x-l}^{x
}= k (μ_{0}I/(4πy))[x/(x^{2}
+ y^{2})^{1/2} - (x - l)/((x - l)^{2}
+ y^{2})^{1/2}].(b) F = LI, F = ∫ _{A}B(r)∙n dA.Flux due to side 1: F _{1} = ∫_{A}B(r)∙k
dA= (μ _{0}I/(4π))∫_{0}^{l}∫_{0}^{l}dx
dy y^{-1}[x/(x^{2}
+ y^{2})^{1/2} - (x - l)/((x - l)^{2}
+ y^{2})^{1/2}]. L = (μ _{0}I/π)∫_{0}^{l}∫_{0}^{l}dx
dy y^{-1}[x/(x^{2}
+ y^{2})^{1/2} - (x - l)/((x - l)^{2}
+ y^{2})^{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)). |

Consider the impedance bridge shown in the figure below. Its purpose is to permit
measurements of an unknown impedance Z_{u} in terms of the fixed
resistance R_{A} and R_{B}, variable resistance R_{S}
and variable capacitance C_{S}. If Z_{u} is a pure
resistance, then C_{S} may be removed from the circuit (shorted out) and
the impedance bridge becomes a simple Wheatstone bridge.

(a) For a purely resistive impedance Z_{u} find the value of R_{S}
for which a balance is obtained (no current on the null detector).

(b) For a complex impedance Z_{u} whose resistive component is R_{u}
and whose reactive component is X_{u}, find the values of R_{S}
and C_{S} for which balance is obtained. Assume Z_{u} has
zero inductance.

(c) Show that for a purely inductive component Z_{u} balance is
not possible.

Solution:

Concepts: AC circuits | |

Reasoning: The circuit shown is a simple, two terminal, AC circuit. | |

Details of the calculation: (a) The circuit in part (a) is equivalent to the circuit shown below. We want V _{b} = V_{a}.V = I _{1}(R_{B} + R_{S}) = I_{2}(R_{A}
+ R_{u}), We want I_{1}R_{B} = I_{2}R_{A},
I_{1}R_{S} = I_{2}R_{u}.R _{S} = (R_{B}/R_{A})R_{u}.(b) The circuit in part (b) is equivalent to the circuit shown below. We want V _{b} = V_{a}.We use the same arguments as in part (a) to show Z _{S} = (R_{B}/R_{A})Z_{u}
= (R_{B}/R_{A})(R_{u} + iX_{u})).Z _{S} = (R_{B}/R_{A})R_{u} + iR_{B}X_{u}/R_{A})
= R_{S} - i/(ωC_{S}). R_{S}
= (R_{B}/R_{A})R_{u}, C_{s} = -R_{A}/(R_{B}X_{u}ω).If Z _{u} = (R_{u} + 1/(iωC_{u}))
then R_{s}
= (R_{B}/R_{A})R_{u},
C_{s} = R_{A}C_{u}/R_{B}.(c) If Z _{u} = iωL_{u},
we need R_{S} - i/(ωC_{S}) = iω(R_{B}/R_{A})L_{u},
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 = 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/mWe 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. |

**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?

** **Solution:

Concepts: Circuits | |

Reasoning: The motor has resistance and inductance. | |

Details 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, ε _{back} = 145 V andI _{peak} = (ε - ε_{back})/R = (240 V - 145 V)/(30 Ω) =
3.2 A.(c) At a time when I = 6.0 A we have ε _{back} = ε - IR = 240 V - (6 A)(30 Ω) = 60 V. |