PHY2049 - Fall 2018 - Exam 2 solutions Problem 1

Solution:

Problem 2

Solution:

Use Kirchhoff's rules (two loops; two junction points):1) Loop equation for the left loop (moving from bottom-left coner clockwise): ε

1− i

1R

1− i

3R

3= 0

2) Loop equation for the left loop (moving from bottom-right coner counterclockwise): ε2− i

2R

2− i

3R

3= 0

3) Current equation for the top junction point: i1+ i

2= i

3

Current equation for the bottom junction is identical to the one above⇒ Three equations with three uknown currents ⇒ Solve for all currents, including i

3

Problem 3

Solution:

The value of the initial charge does not enter the final answer Problem 4

Solution:

Two resistors, the internal of the battery (r=1 ohm) and the external one (R=4 ohm), are connected in series making an equivalent resistance of 5 ohm. With a 10 V battery, this gives a current of 2 A in the circuit. Hence, the power dissipated in the battery is

i(t) = V0Re−t /τ = (Q /C)

Re−t /τ , where τ = RC

P(t) = i(t)2R

P(t) = 12P0 ⇒ (Q /C)2

Re−2t /τ = 1

2(Q /C)2

R

e−2t /τ = 12

−2t / τ = −ln(2)

t = ln(2)2

τ

P = i2r = (2 A)2 (1 ohm) = 4 W

Problem 5

Solution:

To find the direction of the force, use the right-hand rule (and remember that the charge is negative!) Problem 6

Solution:

!F = q!v ×

!B

Circular motion of a charge in uniform magnetic field: R = mvqB

⇒ v = qmRB

⇒ K =12mv2

An use the following conversions: 1 eV = 1.6×10-19 J1 MeV=106 eV

Problem 7

Solution: The force due to electric field is parallel to the field. The force due to magnetic field is perpendicular to the field. For two forces to have a chance to cancel, the two fields must be perpendicular to each other, which is not the case in this problem. Hence, the net force is never zero ins such a configuration of E and B fields. Problem 8

Solution:

First wire with current i1 makes magnetic fields in the place where the loop is placed: B = µ

0

2πi1

R

The force on the top side of the loop with i2: !F =!i

2L×!B = i

2L( ) µ

0

2πi1

a

⎛

⎝⎜

⎞

⎠⎟ upward

The force on the bottom side of the loop with i2: !F =!i

2L×!B = i

2L( ) µ

0

2πi1

a+b

⎛

⎝⎜

⎞

⎠⎟ downward

The forces on left and right sides are zero, since !i

2⊥!B

The net force: i2L( ) µ

0

2πi1

a

⎛

⎝⎜

⎞

⎠⎟− i2L( ) µ

0

2πi1

a+b

⎛

⎝⎜

⎞

⎠⎟

Problem 9

Solution:

B1 is the field due to current i1. B2 is the field due to current i2. At point (x,y)=(a,b)=(1,2), the two fields are equal in magnitude and point in the opposite directions; hence, the net field at the point of interest is zero. To get the direction of magnetic field, use the right-hand rule for wires with a current.

Problem 10

Solution:

There "hair pin" can be viewed with a half-infinite wire at the top, an arc of radius R, and another half-infinite wire.

The point O is

- Distance R away from the edge of the top half-infinite wire: B1 =12

µ0

2πiR

⎛

⎝⎜

⎞

⎠⎟

- At the center point of the arc: B2 =µ0

4πiRΔφ =

µ0

4πiRπ

- Distance R away from the edge of the bottom half-infinite wire: B3 =12

µ0

2πiR

⎛

⎝⎜

⎞

⎠⎟

All three contributing fields are pointing into the page (use right-hand rule for the field direction around a current):

B = B1 + B2 + B3 =µ0

4πiR

(2+π )

Problem 11

Solution:

Problem 12

Solution: Parallel currents in the same direction attract each other; in the opposite directions – repel. The force per unit of length is

Magnetic dipole moment of one electron at a distance z away along the axis of the dipole moment direction is !B =

µ0

2π

!µ

1

z3

Potential energy of the second electron associated with orientation of its magentic dipole moment

in the field of the produced by the first electron is U = −!µ

2⋅!B =

µ0

2πµ 2

z3cosθ.

The difference of potential energies for θ = 0o and θ =180o is ΔU = 2µ

0

2πµ 2

z3

Convert Joules to eV: 1 J = 1.6 ×10-19 eV

F = µ02πi1i2d

Problem 13

Solution:

Problem 14

Solution: The magnetic field at the location of the loop is increasing and so does the flux. According to Lenz’s law, the consequences of the induced current in the loop will be such as to prevent the change of the flux. One of such consequences is the force on the loop DOWN, i.e. to the region of smaller field which would decrease the flux.

Inductance of a solenoid: L = µ0n2ℓ = µ

0

Nℓ

⎛

⎝⎜

⎞

⎠⎟

2

ℓ = µ0

N 2

ℓ, where

N - number of wire turns, ℓ - solenoid length

By winding a 4.6 m wire on a cylinder of diameter 1 cm, one will make N =4.6

π ⋅0.01 turns.

If the wire diameter is 0.5 mm, the length of the solenoid ℓ = N ⋅ (0.0005 m)

Problem 15

Solution: The field of the central loop is into the page inside the loop and out from the page everywhere else; hence, the flux through loops 1 and 3 is out of the page. Since the current in loop 2 increases and the magnetic field everywhere increases, the flux of the magnetic field in loops 2 and 3 increases. Induced EMF in these two loops is, therefore, going to be clockwise, according to Faraday’s Law:

Problem 16

Solution: Equation for inductor’s behavior in a circuit:

Since the switch has been closed for a long time and currents in the circuit do not change anymore, the inductor behaves simply as a conductive wire. The battery “sees” an equivalent resistance of 2 ohm. Hence, the current through the battery is (6 V) / (2 ohm) = 3 A, all of which goes via the inductor. As one opens the switch, the current through inductor cannot change abruptly and remains 3 A. This current can go only trough resistor R2 leading to the difference of potentials on this resistor (3 A) (4 ohm) = 12 V

εinduced = − dΦdt

ε = −L didt

Problem 17

Solution:

Problem 18

Solution:

Problem 19

Solution:

From the law of energy conservation: 12q

02

C+

12Li

02 =

12qx2

C+0; ⇒ q

x= ...

i = i0e−t /τ( )cos(ω 't), where τ= 2L

R

Time for the amplitude to decrease by a factor of two: i = i0e−t /τ( ) = 1

2i

0

⇒ t = ln(2)τ = ln(2) 2LR

If one doubles L and R, t does not change. (Capaciatnce C does not affect the answer.)

imax=ε

max

Z, where Z= R2 + ωd L−

1ωdC

⎛

⎝⎜

⎞

⎠⎟

2

Problem 20

Solution:

irmsUS =εrmsUSX L

=εrmsUSωUS L

=εrmsUS2π fUS L

irmsUK = ...= εrmsUK

2π fUK LirmsUS

irmsUK=εrmsUSεrmsUK

fUKfUS

=110220

5060

≈ 0.42