Exam 2 (070709) solutions

Notes:

1) Unfortunately, the diagrams are missing; please refer to your exam copy

2) Also, the @#$%&*+! software doesn't always properly reproduce the symbols, esp. Greek characters. Please use your native intelligence to figure it out & email me if you have difficulty.

3) RHR = Right Hand Rule

3) These 'solutions' are indications of how to do the calculations; some units and decimal details have been omitted.

Again, contact me if there are problems.

1. The current-carrying loop in the diagram is released from rest in the

uniform magnetic field shown. About which axis does it rotate?

(1) x-axis

(2) y-axis

(3) z-axis

(4) There is no torque on the loop so it does not rotate.

(5) Depends on whether the charge carriers are positive or negative.

The loop's magnetic moment [mm] is into the page [RHR]; the torque is in a direction so as to align the mm with B. Thus rotation is about the x-axis

2. A long, straight wire carrying a current I is bent into a 90-degree

angle at its midpoint as shown in the figure. Approximately what is

the magnetic field at point P, at distance x from the bend and on a

line with the horizontal segment?

(1) μo I /4πx (2) μo I /2πx (3) μo I /√2πx (4) muo √2I /πx (5) none of these

Point P is distance x from half of a long straight wire [the vertical segment]! Thus B = 1/2[µ₀I/(2πx) = [µ₀I/(4πx)

3. An electron moves with velocity 1000 m/s along the x-axis in the (–x) direction. A uniform magnetic field B = 5T is directed in the (+x) direction. The magnitude of the electromagnetic force on the electron, in femto newtons [N ×10− 15 ] is

(1) zero (2) 0.8 (3) 0.4 (4) 1.6 (5) 0.0008

The electron is moving parallel to B, so there is zero magnetic force.

4. A circular loop of conducting wire is placed half in and half out of a

square region of uniform magnetic field directed into the page as shown.

In order to induce a clockwise current in the loop:

(1) move it in the +x direction

(2) move it in the +y direction

(3) move it in the -x direction

(4) move it in the -y direction

(5) increase the magnetic field

An induced current must produce a B field that opposes the flux ) change. Since B is constant, the only way to change the flux is to change the loop area within B. The required (by the question) clockwise loop current will produce B in the same direction as the applied B [RHR], thus we want the flux to decrease. Therefore the loop must be moved out of the applied B = +x direction.

5. The loop in the above question is now relocated to its left so as to be completely immersed in the applied field B which is perpendicular to the plane of the loop. The loop is rotated about a point on one edge at constant speed v, keeping the plane always perpendicular to B. The emf induced in the loop is

(1)zero (2)2πv/B (3)v/(2πB) (4)4πvB (5)vB/(4π)

Moving the loop while completely immersed in B doesn't change the flux, so induced emf = 0

6. A particle of mass 4.3 ×10− 21 kg moves in a uniform magnetic field of

B=2.4T and follows the circular path shown in the diagram. If one revolution is completed in

4.2 ×10− 7 s, what is the particle’s charge, in C?

(1)+2.7 ×10−14

(2)+4.3 ×10−13

(3)+1.5 ×10−15

(4)+1.1 ×10−14

(5)none of these

q = mv/Br; also, v = 2πr/T where T = period

Thus q = 2πrm/BrT = 2πm/BT = 2.7 e-14 C.

7. The work done per revolution (in J) by the B-field on the moving charge in the above situation is:

(1)none of these (2)2πmv2 (3)πB/mq (4)mv 2/2π (5)mv2/4π

B fields do no work when deflecting charged particles! They change the direction, but not the speed; hence no change in KE. Thus no work!

So W = 0 ==> (none of these)

8,An 8.0 mH inductor and a 2.5Ω resistor are wired in series to a 23.6V ideal battery. A switch in the circuit is closed at time t=0, at which time the current is 0. At a time 5.4 msec after the switch is thrown the potential difference in volts across the inductor is:

(1)4.4 (2)19.2 (3)13.0 (4)15.2 (5)7.7

∆V_batt = E = ∆V_R + ∆V_L The current is I = E/R [1- exp(-Rt/L)].

= 23.6/2.5 x [1- exp (2.5 x 5.4 msec)/8mH] = 7.7A

The voltage drop across R = IR = 19.2 V; hence that across L is

23.6 - 19.2= 4.4 V

9. Refer to the previous problem. How much energy in joules is stored in the inductor after the switch has been closed a long time?

(1)0.36 (2)0 (3)0.24 (4)0.012 (5)0.17

U = ½ LI² = ½ (8 mH)(7.7)² = 0.36J

10. The primary of a 3:1step-up transformer is connected to a source and the secondary is connected to a resistor R. The power dissipated by R in this situation is P.

If R is connected directly to the source it will dissipate a power of:

(1)P/9 (2)P/3 (3)P (4)3P (5)9P

V₁/V₂ = N₁/N₂ = 1/3 ==> V₁ = V₂/3.

Power in secondary = P = (V₂)²/R ;

Power in primary P₁ = (V₁)²/R = (V₂/3)²/R = (V₂)²/9R = P/9

11. An RLC series circuit, connected to an applied sinusoidal voltage source of amplitude E, is at resonance. Thus:

(1)the voltage amplitude across Ri sequal to E

(2)the voltage amplitude across R is zero

(3 )the voltage amplitude across C is zero

(4)the voltage amplitude across Lequals E

(5) the applied voltage and current are 90◦outof phasewitheachother

At resonance, X_L + X_C = 0 so impedance Z = R ; thus the entire voltage is across R.

12. Immediately after switchS in the circuit shown is closed, the current through the battery shown is:

(1)Vo/(R1+R2) (2)Vo/R1 (3)Vo/R2 (4)0 (5)Vo(R1+R2)/(R1R2)

When S is closed at time t = 0, the back emf in the inductor prevents any current through it. Thus i = V₀/(R₁ + R₂).

13. For the above circuit, a long time after the switch is closed, the current through the inductor will be approximately

(1) Vo /R1 (2) Vo /(R1 + R2 ) (3) Vo /R2 (4) 0 (5) Vo (R1 + R2 )/(R1 R2 )

After a long time, the back emf = 0, so L behaves like a direct connection and the circuit current bypasses R₂ . The current thus is just V₀/R₁ .

14. The entire circuit shown is immersed in a uniform magnetic field that

is into the page and is decreasing in magnitude at the rate 150 T/s.

The current in the circuit (in amperes) is:

(1) 0.18 (2) 0.22 (3) 0.40 (4) 0.62 (5) none of these

The current due to the battery is CCW around the loop;

thus I = 4V/10Ω = 0.4 A

The induced emf due to dB/dt = (loop area) x dB/dt = 2.16 V and is CW [Lenz's Law] thus partially neutralizing the battery emf.

so the net emf = 2.16 V - 0.4 V = 1.76 V and the current I = V/R = 1.76/10 = 0.176 or (rounding off) 0.18A

15. In the circuit segment shown, the current I = 2.0 mA, flowing from A

to B, and the potential difference, VA − VB is +30V.

At this instant,the charge and polarity of the capacitor C will be:

(1) 1.5 mC, left plate is positive

(2) 1.5 mC, right plate is positive

(3) 0.50 mC, left plate is positive

(4) 0.50 mC, right plate is positive

(5) none of these

Starting at point A which is at +30 V with respect to B, moving toward B, and applying the loop rule, we have

+30 V + ∆V_C - 40 V - IR = 0 ==> ∆V_C - 10 V - (2 mA)(100000) = 0

==>∆V_C - 10 V - 20 V = 0 ==> ∆V_C = 30 V

The charge Q = C ∆V_C = 50 x 30 = 1500 mJ = 1.5 mJ

In order for the voltages to sum up correctly, the left plate of C must be positive.

16.The figure shows a simple RC circuit consisting of a 10μF capacitor in

series with a resistor. The capacitor is externally charged to a potential

difference of 100 V while the switch is open. At time t = 0, the switch is

closed. 2.0 s later, the potential difference between the capacitor plates

is 37 V. Calculate the electric potential energy stored in the capacitor

before the switch was closed.

(1) 0.05 J (2) 0.01 J (3) 0.02 J (4) 0.03 J (5) 0.04 J

U = ½ CV² = ½ (10µF) (100)² = 50000 µJ = .05J

17. Refer to the information above. Determine the potential drop across the resistor R at t = 2.0 s (i.e., two seconds after

the switch is closed).

(1) 37 V (2) zero (3) 63 V (4) 87 V (5) 100 V

∆V_R = ∆V_C = 37 V

18. Refer to the information above. Determine the numerical value of the resistance R.

(1) 2 × 10 5 Ω (2) 1.0 × 10 5 Ω (3) 5.0 × 10 5 Ω (4) 1.0 × 10 6 Ω (5) 2.5 × 10 6 Ω

The capacitor's charge q when discharging is q = Q exp (-t/RC) where Q = initial charge. From #16 (above), at t=2.0s,

(10µF) (37 V) = (10µF)(100 V) exp (-2/RC)

37 = 100 exp (-2/RC) ==> 0.37 = exp (-2/RC)

==> 1/0.37 = 2.7 = exp (2/RC)

ln (2.7) =0.993 = 2/RC ==> R = 2/0.994 C = 2/(0.994e-5)

==> R = 2e5Ω

19. In the circuit shown at right, R = R = 10kΩ, C = 200μF, and

E =40 V. When S is closed, the time constant for charging C is (in seconds): R

(1) 2 (2) 4 (3) 0.5 (4) 1 (5) none of these

The capacitive time constant is RC, where R is the charging circuit resistance. When S is closed, the charging current passes through the upper R only, , so RC = (10000) (200 µF) = 2 s

20. For the above circuit, when S is open, the time constant for discharge of capacitor C (in seconds) is

(1) 4 (2) 2 (3) 0.5 (4) 1 (5) none of these

When S is open, C discharges through both resistors, so the circuit resistance = 20,000Ω and RC is thus doubled, to 4 s.