PHY2049Exam2solutions–Fall2016

Solution:Generalstrategy:Findtworesistors,onepairatatime,thatareconnectedeitherinSERIESorinPARALLEL;replacethesetworesistorswithoneofanequivalentresistance.Nowyouhaveacircuitisoneresistorless.Repeattheprocedureuntilyouhaveoneequivalentresistanceleft–thisistheanswer.

Solution:Generalstrategy:1)Definecurrentsineachsegmentofelementsconnectedinseries(directionscanbearbitrary).Seehowmanyunknowncurrentsdoyouget(threeinthiscircuit).So,weneedthreeequationstofindthreeunknowncurrents.Countnumberofloops(twointhiscircuit),eachofwhichwillgiveyoualoopequationforsumofpotentialdifferences.Theremainingequation(thirdequationinthisproblem)willcomefromthejunctionruleforcurrents.

2)Foreachloop,writedownequationsforpotentials:“walk”alongeachloop(inanydirection)andsumalldifferencesofpotentialsasyoujumpoverelementsofacircuit;whenyoucomebacktothestartpoint,thesummustbezero.-leftloop(walkingclockwisefromtheleft-bottomcorner):ε − i1R− i2r − 2ε − i1R = 0 -rightloop(walkingclockwisefrompointb):2ε + i2r − i3R−ε = 0 3)Foreachjunctionofwires,writedownanequationforcurrents:sumofincomingcurrents=sumofoutgoingcurrents.Discardequationsthatarenotindependent.-Pointa: i1 = i2 + i3 Inthisproblem,wecanstop:wealreadyhavethreeequationsforthethreeunknown.Forcuriousminds:Pointbgivetheexactsameequationforcurrentsaspointa,i.e. i2 + i3 = i1 .So,wehaveonlyoneindependentequationforcurrents.4)Solvethethreeequations:ε − i1R− i2r − 2ε − i1R = 0 2ε + i2r − i3R−ε = 0 i1 = i2 + i3

Solution:Powerdissipatedbytheresistor:P = i2R InadischargingRCcircuit:

i(t) = q0

τe−t/τ , where τ = RC

Therefore:

P(t) = i(t)2R = q02

τ 2Re−2t/τ

P(t)P(0)

= e−2t/τ

12= e−2t/τ

t = ln22τ

Solution:

Power dissipated by the resistor r inside the battery: P = i2r,

where i is a current in the circuit, i = εR+ r

Solution:Usetherighthandrule(donotforgetthatthechargeisnegative!)

Solution:Considerdirectionofforcesoneachofthefourwiresegmentsmakingthesquareloop.Leftandrightwirescarrythecurrentparalleltothemagneticfield:forceiszero.Thetopwire(usetherighthandrule):theforceisoutofpage.Thebottomwire:theforceisintothepage.Hence,theloop“wants”toturnaroundthex-axis.

Solution:In the gap with the electric field, ions are accelerated and gain energy: K f −Ki =Ui −Uf

mv2

2= qV ⇒ v = 2qV

mIn the space with the magentic field, ions follow a circular path of radius r:

m v2

r= qvB ⇒ r = mv

qB=mqB

2qVm

~ VB

Solution:Potential energy of a magnetic dipole in a magnetic field: U = −

!µ ⋅!B = −µBcosθ

Torque of a magnetic dipole in a magnetic field: !τ =!µ ×!B = µBsinθ

Therefore: !τ

U= − tanθ

(Keep in mind that an angle between two vectors is between 0! and 180!)

Solution:

Magnetic field made by the wire with current i1 =12A: B = µ0

2πi1r

Consider forces on each side of the loop with current i2 = 2A

The left segment of length b = 8cm, distance a = 4cm away from the wire: FL = i2!b ×!B(a) = i2b

µ0

2πi1a

.

The right segment of length b = 8cm, distance 2a = 2 ⋅ 4cm away from the wire: FR = i2!b ×!B(2a) = i2b

µ0

2πi12a

.

These two forces are in opposite directions, so the magnitude of the net force along the x-axis is FL −FR =µ0

2πi1i22a

b

The forces on the top and bottom part of the loop are same in magnitude (by symmetry) and opposite; so their sum is zero.

Solution:

Quadrant 2: The magnetic field made by the wire with current i1 is out of the page B = µ0

2πi1r=µ0

2πi1y

⎛

⎝⎜

⎞

⎠⎟;

with current i2 –– into the page B = µ0

2πi2r=µ0

2πi2x

⎛

⎝⎜

⎞

⎠⎟.

So the two fields would cancel each other, when their magnitudes are the same: µ0

2πi1y=µ0

2πi2x

,

i.e. along a straight line, given by y = i1i2x

Quadrant 2: The magnetic field made by the wire with current i1 is out of the page; with current i2 ––out the page. So the two fields cannot cancel each other.Quadrant 3: Similar to Quadrant 1Quadrant 4: Similar to Quadrant 2

Solution:The figure represents an infinitely long wire with current i and a circular loop, also with current i.

The magnetic field made by the current in the straight wire at the center of the circle (i.e. at a distance d

from the wire) is B = µ0

2πid

and into the page.

The magnetic field made by the current in the loop at the loop center is B = µ0

4πid

(2π ) and out of the page.

The magnitude of the net field is the difference between the two, i.e. µ0

4πid

(2π ) – µ0

2πid

= µ0

2πidπ −1( )

Solution:

Magnetic field inside a solenoid of length ℓ with density of windings n = Nℓ

(N is number of windings)

and carrying current i is B = µ0in = µ0iNℓ

.

Hence, N =Bℓµ0i

. The information on the radius of the solenoid is irrelevant.

Solution:

The inductance of the original solenoid is L = µ0n2Aℓ = µ0

Nℓ

⎛

⎝⎜

⎞

⎠⎟

2

πr2( )ℓ = µ0πN 2r2

ℓ.

The inductance of the new solenoid: Lnew = µ0π(3N )2 r2

2ℓ=

92L

Solution:

The induced EMF in an inductor is ε = −L didt

,

where didt

at t =1 ms can be read off the graph: didt=

2A−8A2ms

= −3000 As

Therefore: ε = 900 VPotential on the right is higher than on the left, therefore Vright −Vleft = 900 V

Solution:The magnetic field produced by the central loop in places where loops 1 and 3 is out of the page. The magnetic field is increasing.

Therefore, the magnetic field flux in loops 1 and 3 is out of the page and increasing, i.e. dΦdt

is also out of the page.

Induced EMF in loops 1 and 2 is then ε = − dΦdt

, i.e. clockwise, which will result in clockwise currents (in both loops).

(Align your right hand thumb with the direction of − dΦdt

and curl your fingers to get the direction of induced EMF.)

Solution:A coil of area A with N turns rotating with angular frequency ω = 2π f in magnetic filed Bwill generate an induced EMF ε(t) =εmax sin(ωt), where εmax =ANBω.

If the coil has resistance R, the current is then i(t) =ε(t)R

=εmax

Rsin(ωt) = imax sin(ωt), where imax =

ANBωR

.

RMS value of current is irms =imax

2.

Solution:

The energy stored in an LC circuit is conserved. Hence: q2

2C+

12i2L = qmax

2

2C.

(At the time when the charge is maximum, the current must be zero)

Solution:

The decay time constant in an RLC circuit is τ= 2LR

.

If both L and R are doubled, the decay time does not change.(The information on capacitance C is irrelevant.)

Solution:The amplitude of current in an RLC circuit driven by an external EMF is

imax =εmax

X, where X = R2 +

1ωdC

−ωdL⎛

⎝⎜

⎞

⎠⎟

2

Solution:If a capasitor C is connected to an external EMF of freqency f ,RMS value of current in the circuit is

irms =εrms

XC

, where XC =1

ωdC=

12π fC

irmsUS =εrms

US 2π f USCirmsUK =εrms

UK 2π f UKC

Hence: irmsUS

irmsUK =

εrmsUS f US

εrmsUK f UK

= 0.6