October 31, 2008
PHY2054 Discussion-Fall ‘08
Quiz 7 (Chapter 19.7-20.8)
Name:
UFID:
*1. (5pts) Four long, parallel wires carry equal currents of 2.00 A. The wires are directed
perpendicular to the sheet and they pass the corners of a square with edge of length 0.500 m. The
direction of the current is into the sheet for wire A and B and out of the sheet for wire C and D.
Calculate the magnitude and direction of the magnetic field at point P, located at the center of the
square.
According to the 2nd right hand rule, the magnetic fields due to the current
A and D are directed to wire B, and the field due to the current B and C
are directed toward D. Taking +x direction to the right and +y upward, we
have
BA = BD = (-μ0I/(2πr)cos45°, - μ0I/(2πr)sin45°)
BB = BC = (+μ0I/(2πr)cos45°, - μ0I/(2πr)sin45°)
The net magnetic field is the sum of the field due to each current. Therefore
B = BA+BB+BC+BD = (0, -4 μ0I/(2πr)sin45°)
= (0,-4×4π×10-7×2/(2π×√(2×0.252)) = (0, -3.20×10-6 )
Thus the field has a magnitude of 3.20 μT and directed downward in the plane of the page.
*2. (5pts) A square loop with edge length of 0.800 m has a total series resistance of 0.400 Ω. It is
placed in a uniform 0.500-T magnetic field directed perpendicular out of the sheet. The loop is
pulled as shown until the separation between points A and B is 0.400 m. If this process takes 0.200 s,
what is the magnitude and direction of the average current generated in the loop?
The initial area of the square loop is
Ai = l2 = 0.82 = 0.64 m2
We calculate the final area of the rhombus by subdividing it into four
triangles:
Af = 4(1/2)bh = 4×0.5×0.2×√(0.82-0.22) = 0.310 m2
Combining Faraday’s law and Ohm’s law, the current flows induced in the loop is given by
I = Є/R = ΔΦ/R = (1/R)BΔA/Δt = (1/0.4)×0.5×(0.64-0.31) /0.2 = 2.06 A
Since the external flux out of the sheet is decreasing, the induced flux is out of the sheet. The current
creates this induced flux is counterclockwise.
Constants
μ0 = 4π×10-7 Tm/A
me = 9.11×10-31 kg
e = 1.6×10-19 C
**3. (5pts) A horizontal wire is free to slide without friction on the vertical rails of the conducting
frame. The wire has a mass of 0.500 kg and a length of 0.400 m, and the resistance of the circuit is
3.00 mΩ. The circuit is placed in a uniform magnetic field perpendicular to the frame. If the terminal
speed of the wire is 6.00 m/s, what is the strength of the magnetic field?
When the wire falls, the current is induced by the motional emf. The
induced current is given by
I = Є/R = Blv/R
Since the current is proportional to the velocity, as the velocity
increases, the magnetic force increases and it eventually cancels the
force of gravity. Applying the equilibrium condition to the wire, we
express the magnetic field in terms of the terminal velocity:
Fm-Fg =0 ⇒ IBl-mg = 0 ⇒ (Blv/R)Bl-mg = 0
B = √(mgR/(l2v)) = √[0.5×9.8×3×10-3/(0.42×6)] = 0.124 T
***4. (5pts) A 6.00 V battery is connected in series with two identical 1.00-Ω resistors and an
inductor. If the two resistors are connected in series, it takes the current 5.00 ms to reach 2.00 A after
the switch is closed. How long does it take the current to reach 2.00 A if the resistors are connected
in parallel?
The maximum current in the left circuit is
I0 = ΔV/Req = ΔV/(2R) = 6/2 = 3 A
Using the formula for the current in RL circuit,
we
calculate
the
time constant
and the
inductance
I(t) = I0(1-e ) ⇒ 2 = 3[1-exp(-5×10 /τ)]
-t/τ
-3
⇒ exp(-5×10-3/τ) = 1/3 ⇒ (-5×10-3/τ) = ln(1/3) ⇒ τ = -5×10-3/ ln(1/3) = 4.55×10-3s
The inductance of the coil and the time constant for the right circuit are
τ = L/(2R) ⇒ L = 2Rτ = 2×1×4.55×10-3 = 9.10×10-3 H
τ’ = L/Req’ = L/(R/2) = 2×9.10×10-3/1 = 18.2×10-3 s
After getting the maximum current in the right circuit, we plug the maximum current and the time
constant into the formula for RL circuit. We have
I0’ = ΔV/Req’ = 6/(1/2) = 12.0 A
I(t) = I0’(1-e-t/τ’) ⇒ 2 = 12[1-exp(-t/(18.2×10-3))]
⇒ exp(-t/(18.2×10-3) = 5/6 ⇒ -t/(18.2×10-3) = ln(5/6) ⇒ t = -18.2×10-3× ln(5/6) = 3.32 ×10-3 s