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PHY 2049 Spring 2013 Exam 2 solutions

Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 A large appliance consumes 10 kW of electrical power from a utility company that delivers that power with an EMF of 400 V. What is the internal resistance of the appliance? (1) 16 Ω (2) 25 Ω (3) 40 Ω (4) 0.04 Ω (5) 16,000 Ω

P = ε 2

R

⇒ R = ε 2

P=400( )210000

= 16Ω

Problem 2 Two cylindrical wires are made from copper. The second wire has twice the diameter and twice the length of the first wire. What is the ratio of the resistance of the second wire to that of the first? (1) 1/2 (2) 2 (3) 1/4 (4) 4 (5) 1

R1 = ρ L1A1

R2 = ρ L2A2

R2R1

= L2L1

A1A2

= L2L1r12

r22 =

21122

= 12

Problem 3 Two light bulbs each with a resistance of 250 Ω can be connected in parallel or in series to a source of EMF with ε = 120 V. What is the ratio of the brightness of one of the light bulbs when it is in the parallel circuit to that of one of the light bulbs when it is in the series circuit? (The two light bulbs have the same brightness in each circuit since they are identical.) (1) 4 (2) 2 (3) 16 (4) 1/2 (5) 1/4


With the light bulbs in series, the current through each light bulb is:

i = ε2R

The power dissipated by one bulb (its brightness) is:

Ps = i2R = ε 2

4R

With the light bulbs in parallel, the current from the EMF source is:

i = εR / 2( ) =

2εR (resistors in parallel)

The current through each bulb is half of the total current (evenly split into each parallel path). So the power dissipated by one bulb (its brightness) is:

Pp =i2

⎛⎝⎜

⎞⎠⎟2

R = 2ε2R

⎛⎝⎜

⎞⎠⎟2

R = ε 2

R

So the ratio of the brightness of one bulb in the parallel circuit to one in the series circuit is: PpPs

=ε 2 / R( )ε 2 / 4R( ) = 4

Problem 4 Find the equivalent resistance of the circuit connected to the source of EMF as shown in the figure if every resistor has the same resistance 60Ω.

(1) 80 Ω (2) 150 Ω (3) 120 Ω (4) 15 Ω (5) 60 Ω

The 3 resistors on the right, R2-‐R4, are all in parallel with equivalent resistance R/3. These are in series with R1, so the total equivalent resistance is R+R/3 = 4R/3 = 80 Ω


Problem 5 How long does it take to charge an initially uncharged capacitor to half of its maximum charge of 24μC if it is connected in series with a 30 kΩ (or 15 kΩ) resistor to a 12 V EMF source? (1) 0.04 s (2) 0.06 s (3) 0.03 s (4) 0.02 s (5) 0.4 s

This is an RC circuit, so the charge on the capacitor follows an exponential plus a constant. Boundary conditions imply that: q t( ) = qm 1− e− t /RC( ) The maximum value of the charge, qm, is given when the capacitor is fully charged to the voltage of the EMF source (current goes to zero, no voltage across resistor). This tells us: qm = Cε , or that C = qm / ε When the capacitor is half charged: qm2

= qm 1− e− t /RC( )

12= 1− e− t /RC

12= e− t /RC

ln2 = tRC

t = RC ln2 = Rqmεln2

which is 0.04s or 0.02s depending on R. Problem 6 In the circuit shown, the EMF source maintains a potential difference of ε = 6 V and the resistances are R1 = 10Ω (or 20Ω), R2 = 20Ω (or 10Ω), and R3 = 30Ω. If the electric potential at point c is Vc = −3 V, what is the electric potential Vb at point b?

(1) +1 V (2) +4 V (3) −1 V (4) +2 V (5) 6 V


Note that Vac = ε Consider the path ac:

ε = iR1 + iR2 ⇒ i = εR1 + R2

iR1 = ε R1R1 + R2

iR2 = ε R2R1 + R2

⇒Vb =Vc + iR2

Problem 7 Find the magnitude of the current in amps through resistor R3 in the circuit shown if ε1 = ε2 = 9 V and R1 = R2 = 40Ω (or 80 Ω) and R3 = 10 Ω (or 20 Ω).

(1) 0.3 (2) 0.15 (3) 0.1 (4) 0.05 (5) 0

Since R1 = R2 and ε1 = ε2, the currents through those two branches must be the same by symmetry, and equal to half the current through R3. Applying the loop rule to either branch gives us: ε1 − i1R1 − i3R3 = 0

ε1 −i32

⎛⎝⎜

⎞⎠⎟ R1 − i3R3 = 0

⇒ i3 =ε1

R12+ R3

which gives 0.3A or 0.15A depending on the inputs.


Problem 8 Determine the drift velocity vd of free electrons in a copper wire carrying a current of 1 A, if the wire has a cylindrical cross section of radius r=0.005 m. The number density of free electrons in copper is n=8.5 x 1028 m−3. The current density is related to the drift velocity by j = n e vd. (1) 9.4 x 10−7 m/s (2) 7.4 x 10−11 m/s (3) 1.1 x 106 m/s (4) 1.5 x 10−25 m/s (5) 3 x 108 m/s

vd =jne

=i / A( )ne

= iπr2ne

= 9.4 ×10−7 m/s

Problem 9 Consider a charged particle of mass 2 x 10−9 kg which moves in a region of zero electric field and a constant magnetic field of magnitude 5 Tesla and directed straight up from the paper. Suppose the particle moves at a constant speed of 1000 m/s counter-‐clockwise around a circle of radius 3 m. What is the particle's charge in Coulombs? (1) −1.3 x 10−7 (2) +1.3 x 10−7 (3) −8.4 x 10−7 (4) +8.7 x 10−7 (5) +2.1 x 10−8

Because the particle is moving counter-‐clockwise (as viewed from the tip of the magnetic field vector) it must have a negative charge. To find the magnitude q of the charge, recall that the angular frequency is ωB = qB/m. We set this equal to its speed over the radius of the circle so q = mv/BR.


Problem 10 Suppose a charged particle moves with velocity (2 i − 3 k ) m/s in a magnetic field of 8 j nano−Tesla. What must be the electric field (in nano−Volt/meter) in order for the particle to move freely with constant velocity? (1) −24 i − 16 k (2) +24 i + 16 k (3) −24 i + 16 k (4) +24 i − 16 k (5) Must know the charge to answer. For the electric force to cancel the magnetic one we must have

E = −v ×B . Recall how the unit vectors cross into one another: i × j = k

j× k = i

k × i = j

Problem 11 Consider an unknown conductor in the shape of a rectangular bar along the y axis. Suppose the conductor carries a current directed along the positive y axis. A magnetic field is applied along the positive z axis, and the resulting Hall voltage is measured to be positive from the −x side of the bar to the +x side. What is the sign of the conductor's charge carriers and the direction of their average motion? (1) q > 0 and v_y > 0 (2) q < 0 and v_y > 0 (3) q > 0 and v_y < 0 (4) q < 0 and v_y < 0 (5) Insufficient information

Because the current moves in the +y direction, charge carries with q > 0 have vy >0, whereas those with q < 0 have vy < 0. In both case the magnetic force qv ×B points in the +x direction, which makes the charge carriers accumulate on the +x side. Because the Hall potential is positive from –x to +x, the charge carriers must have q > 0, which implies vy > 0.


Problem 12 100 turns of wire are wound around the frame of a door of height 2 meters and width 1 meter. The wire carries a current of 5 Amps up its hinge. Suppose a constant magnetic field of magnitude .003 Tesla is applied parallel to the door's hinge. What is the magnitude of the torque (in N−m) on the door, and which way does it make the door turn as viewed from above?

(1) 0, the door doesn't turn (2) 3, clockwise (3) 3, counter−clockwise (4) 0.03, clockwise (5) 0.03, counter−clockwise

This problem is designed to catch unwary students who compute magnitudes without worrying about directions. The door forms a magnetic dipole of magnitude 1000 A-‐m2, directed into the page. This means the MAGNETIC COMPONENT of the total torque has magnitude 3 N-‐m and points away from the hinge, parallel to the door. Such a torque would tend to cause the door to rotate about the same axis. Of course it is NOT FREE to rotate that way: contact forces from the hinge cancel the magnetic torque so that the total torque is zero and the door doesn't turn. MORAL: vectors have DIRECTIONS as well as magnitudes and these directions matter.


Problem 13 What is the magnetic field (in nano−Tesla) at a point 4 meters along the positive x axis by a wire segment of length 3 meters which carries a current of 5 Amps along the positive z axis?

(1) +75 j (2) −75 j (3) +13 j (4) −13 j (5) +6 j If we measure all lengths in meters then the Biot-‐Savart law gives

B = µ0i4π

d0

3

∫ xk × (4i − zk) / (16 + z2 )32 = µ0i

4 pi4 j d

0

3

∫ z / (16 + z2 )3/2

Now use the integral given on the front sheet (for a=3 and b=4) to conclude

B = µ0i4π

320j = 75 nTj

Problem 14 A cable of radius R carries a current density directed along its axis whose magnitude grows linearly with the distance from the cable's central axis. If the total current in the cable is i0, what is the magnitude of the magnetic field a distance r = R/2 from the central axis? (1) μ0 i0/8π R (2) μ0 i0/6π R (3) μ0 i0/4π R (4) μ0 i0/2π R (5) μ0 i0/π R

As promised in lecture, this is the problem of a nonlinear current density similar to the electric field example of Exam 1. Use Ampere's law. By symmetry, the magnetic field is directed axially around the wire, and its magnitude depends only on the distance from the central axis. We can best exploit this by choosing the surfaces to be circles of radius r, perpendicular to the central axis and centered on it. The boundary of such a circle is its circumference so d∫ s ⋅B = 2πrB(r)


By Ampere's law this must be µ0 2π d0

r

∫ ssj(s)

so B(r) = µ0 / r d0

r

∫ ssj(s)

Now use the fact that j(s) = j0 s. For r < R we have B(r) = µ0 j0r2 / 3

The total current is i0 = 2π j0R3 / 3 , hence B(r) = µ0i0

2πr2 / R3

For r = R/2 we have B = µ0i08πR

Problem 15 Consider three parallel wires which are arranged from top to bottom carrying currents in the same direction. Suppose the top wire carries four times the current of the bottom wire and the middle wire carries twice the current of the bottom wire. If the distance between the top wire and the middle one is d, what must be the distance between the middle wire and the bottom one in order that the force on the middle wire is zero? (1) d/4 (2) d/2 (3) d/8 (4) d/16 (5) d/32

Recall that parallel wires carrying current in the same direction attract one another, whereas those with opposite currents repel one another. Hence the top wire pulls the middle wire up, whereas the bottom wire pulls it down. In each case the magnitude of the force goes like the current in the wire (top or bottom) over its distance from the middle wire. For the forces to balance we must have itop / dtop = ibot / dbot , which implies dbot = dtopibot / itop = d / 4

Problem 16 Consider an infinite solenoid which is made by winding 6000 turns per meter of a wire which carries 3 Amps. Inside the solenoid there is a loop of radius 0.1 m which carries a current of .3 Amps. How much energy is required to flip the loop from its lowest energy orientation to its highest energy orientation? (1) 0.43 mJ (2) 0.034 mJ (3) 0.21 mJ (4) 0.071 μJ (5) 0 J Recall that the magnetic field of a solenoid has magnitude Bsol = µ0isolNsol ≈ 0.0226 T

The magnetic dipole moment of the loop is µ = πR2i ≈ 0.00942 A-m2 The energy of a magnetic dipole in a magnetic field is −µ ⋅B . Hence the maximum energy difference is 2µB ≈ 0.000426 J


Problem 17 A circular conducting loop is immersed in a uniform magnetic field B pointing out of the page with a constant magnitude of 5 T. If the loop suddenly contracts from its initial radius of 0.2 m with a radial velocity dr/dt = −0.2 m/s, what is the induced current and its direction immediately after it starts contracting if the loop has a constant resistance of 0.1Ω? A clockwise direction is considered positive, counter-‐clockwise is negative.

(1) −12.5 A (2) +12.5 A (3) −6.3 A (4) +6.3 A (5) 0

If the loop shrinks, the magnetic flux decreases. Lenz’s Law tells us that the direction of the induced EMF and current will oppose this change, so we need an induced field also pointing out of the page. By a right-‐hand rule, this means that the current must travel counter-‐clockwise (negative).

The radius of the loop can be written: r t( ) = r0 + at , where a =drdt

= −0.2

The magnitude of the induced EMF is given by Faraday’s Law:

ε = dΦB

dt= ddt

BA( ) = B dAdt

= Bπ ddt

r0 + at( )2

ε = Bπ 2r0a initially for t = 0

The current is therefore: i = εR= −12.5 A using the sign for the direction.

Problem 18 A square metal loop of side length l =1.5 m is pulled out of a uniform magnetic field with a velocity of magnitude v=2 m/s. The magnetic field has a strength of B=0.5 T and is directed out of the page and perpendicular to the surface of the loop. One side of the square is aligned with the edge of the field region when the pulling first starts. What is the magnitude of the force that must be applied to pull the loop with this velocity if the resistance of the loop is 0.2Ω?


(1) 5.6 N (2) 1.1 N (3) 1.5 N (4) 7.5 N (5) 3.8 N

The induced EMF in the loop is ε = vLB from Faraday’s law, since the amount of magnetic flux contained by the square is decreasing. Since the flux is decreasing, Lenz’s Law says that the direction of the induced EMF and current will be to oppose that decrease, so the induced magnetic field should point out of the page also. This requires a counter-‐clockwise direction. The

magnitude of the current is i = vLBR

With a counter-‐clockwise current, the current travels down the left edge of the square. This leads to a force pointing to the left: F = iL×B

The magnitude of the force is F = L2B2 vR= 5.6 N

Note that the forces on the top and bottom legs cancel out, so only the force to the left needs to be overcome.

Problem 19 The current through an inductor with inductance L=0.05 H is shown by the graph, with the direction from left to right through the inductor as shown. What is the EMF across the inductor (VL= Vright − Vleft), including sign, at t = 4 ms?

(1) 33 V (2) −670 V (3) −70 V (4) 0.1 V (5) −25 V

The induced EMF on an inductor is given by: ε = −L didt

At t=4ms, didt

=5 − 7( )

0.005 − 0.002( ) = − 20.003

= −667 As

So the EMF is VL = −0.05 −667( ) = 33 V


Problem 20 In the figure, a uniform magnetic field points out of the page and varies in time according to B(t) = 0.1 t + 0.01 t2, with t measured in seconds and B in Tesla. What is the magnitude of

E ⋅ds∫ when t=2 s along a closed path in the plane of the page

that traces an equilateral triangle of side length s = 1 cm as shown in the figure?

(1) 6.1x 10−6 V (2) 1.0x 10−5 V (3) 0.14 V (4) 8.7x 10−7 V (5) 0

ε =E ⋅ds∫ = dΦB

dt= A dB

dtE ⋅ds∫ = 1

2s 3

2s

⎛

⎝⎜⎞

⎠⎟0.1+ 2 0.01( )t( )

E ⋅ds∫ = 6.1×10−6 V

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