March 17, 2009
PHY2054 Discussion-Spring ‘09
Quiz 6 (Chapter 19.7-20.4)
Name:
UFID:
**1. (2.5pts) A long straight wire carrying a current I1 =20.0 A lies in the same plane of a rectangular
loop carrying a current I2 = 5.00 A as shown below. The dimensions shown are c = 0.100 m, a =
0.200 m and l = 0.500 m. What is the magnitude and direction of the net force exerted by the
magnetic field due to the straight wire on the loop?
Two straight currents in the same direction attract each other and those in
opposite directions repel each other, thus the force on the left side of the loop is
to the right and that on the right side of the loop is to the left. Taking the
positive direction to the right, the net force on the loop is
F = Fl – Fr = μ0I1I2l/(2πc)- μ0I1I2l/[2π(c+a)]
= 4π×10-7×20×5×0.5/(2π)×[1/0.1-1/0.3] = 6.67×10-5 N to the right
*2. (2.5pts) A rectangular box with dimensions 8 cm × 10 cm × 9 cm is immersed in a uniform
magnetic field with components Bx = 0.20 T, By = 0.80 T and Bz = -0.40 T. Calculate the flux
through the bottom face (the face in the xz-plane).
The normal to the bottom face is to the negative y direction. Thus the flux
is
ΦB = -ByA = -0.8×0.08×0.1= -6.4×10-3 Wb
Constants & Formulas
μ0 = 4π×10-7 Tm/A
Magnetic Field Inside a Solenoid: B = μ0 nI
Magnetic Force on a Current: F = IBlsinθ
Magnetic Flux: ΦB = BAcosθ
Magnetic Force between Two Parallel Wires:
Faraday’s Law: ε = -N(ΔΦ/Δt)
F/l = μ0I1I2/(2πd)
Motional Emf: |ε| = Blv
Kirchhoff’s Rules: Iin = Iout, ΣΔV = 0
*3. (2.5pts) A circular single-turn loop of a radius 10.0 cm and resistance 2.00 mΩ is placed at the
center of a solenoid. The solenoid is 75.0 cm long, has a circular cross section of radius 5.00 cm and
is wound with 1 500 turns of wire. If the current in the solenoid reduces 2.00 A to zero in 0.400 s,
what is the magnitude and direction of the average induced current in the loop?
The magnetic field outside of the solenoid is zero. Therefore, the initial flux
through the loop is
Φi = BiAsolenoid = μ0 nI Asolenoid = 4π×10-7×(1500/0.75)×2×π×0.052 = 3.95×10-5
Wb
Since the final magnetic flux is zero, the current in the loop is
I = 1/R× |ΔΦ/Δt| = 1/(2×10-3)× 3.95×10-5/0.4 = 4.94×10-2 A
When the current is decreasing, the downward field is decreasing. Thus the induced field is also
downward. The induced current that produces a downward field is clockwise as viewed from the top.
***4. (2.5pts) A horizontal wire is free to slide on the vertical rails of a conducting frame without
friction. The wire has a mass of 0.800 kg and length of 0.400 m. The resistance of the circuit is
provided by two resistors, R1 = 18.0 mΩ and R2 = 2.00 mΩ. The battery provides a potential
difference of 1.50 V. If the circuit is in a uniform 0.100-T magnetic field perpendicular to the frame,
what is the terminal speed of the wire?
The net force exerted on the wire is zero when it moves at the terminal speed.
Thus we have
ΣF = Fm-Fg = 0 ⇒ BIl-mg = 0
where the current I in the wire is determined by Kirchhoff’s rules:
vBl-I1R1+ΔV = 0 (Upper loop), vBl-I2R2 = 0 (Lower loop) & I = I1+I2
(Junction) ⇒ I = vBl/R+ΔV/R1, where R = (1/R1+1/R2)-1 = 1.8 mΩ
Plugging this into the first equation, the terminal velocity is
B( vBl/R+ΔV/R1)l-mg = 0 ⇒ v = mgR/(Bl)2-ΔVR/(BlR1)
= 0.8×9.8×1.8×10-3/(0.1×0.4)2 -1.5×1.8×10-3/(0.1×0.4×18×10-3) = 5.07 m/s