Download Physics 203 Sample Exam 1

Physics 203 Sample Exam 1
Gothard Grey
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Mass of proton = 1.67E-27 kg
Mass of alpha = 4 x mass of proton
Charge of alpha = +4e
+ my 3 x 5 card which is on line.
Name _________________
Multiple choice, 3 points each
[1] An electron a proton and an alpha particle enter a uniform magnetic field, all moving
at the same velocity. The particle with the largest radius for its trajectory is the
(a) electron.
(b) proton.
(c) alpha particle.
(d) all the same.
[2] For the particles in question 1, which particle curves in the opposite direction from
the other two upon entering the field?
(a) electron.
(b) proton.
(c) alpha particle.
(d) all the same.
[3] As you move away from a long, straight wire carrying an electric current, the strength
of the magnetic field varies as
(a) 1/r2
(b) 1/r
(c) ln(r)
(d) r
[4] A magnetic field exerts a force on all the following EXCEPT
(a) a current carrying wire (b) a moving electric charge
(c) a stationary electric charge
(d) a magnet
[5] Which of the following will not drive an electric current in a conductive loop?
(a) magnetic flux through the loop
(b) a potential difference along the loop
(c) an electric field around the loop (d) a changing magnetic flux through the loop
[6] The presence of a ferromagnetic core in the center of a coil
(a) concentrates the magnetic field. (b) spreads the magnetic field lines out.
(c) only serves to support the coil. (d) is necessary to produce a magnetic field.
[7] A light wave is made of
(a) constant electric and magnetic fields.
(b) oscillating electric and magnetic fields in the same direction.
(c) electric and magnetic fields at various angles.
(d) oscillating electric and magnetic fields at right angles.
[8] Magnetic fields can be produced by
(a) electric currents
(b) changing electric flux
(c) both a and b
(d) none of these
[9] The circuit in the diagram at right will allow
what signals to pass?
(a) low frequencies
(b) high frequencies
(c) DC signals
(d) no signals.
[10] The energy stored in an electric field is proportional to
(a) 1/E2
(b)1/E
(c) E
(d)E2
Choose one of problems A or B to do. Show all work. If you have extra time, you
may work on the other for extra credit. 5 points per part.
[A] Two solenoids are next to each other, oriented end to end along the same axis.
Solenoid A is 20 cm long, 4 cm. in diameter and has 200 turns.
Solenoid B has 5 turns per cm., is 15 cm long and 5 cm. in diameter.
(a) Calculate the inductance of solenoid B.
L = NBA/I = N2oR2/l = (5x15)2 (4E-7 Tm/A)(2.5E-2)2 m2/0.15m
= 9.25E-5 Tm2/A
(b) Given a current of 4sin(10t) in solenoid A, calculate the flux in solenoid B. You may
assume no leakage of field from between the solenoids .
Flux in B = BA for coil B = (onAI)(rA2) since B field confined to area of
coil A (or spreads out to give same flux in coil B.
Flux = (4E-7 Tm/A)(1000 turns/m)(4sin(10t) A)(.02m)2
= 6.32E-6 Tm2
(c) Calculate the induced voltage in solenoid B given a magnetic field of 4sin(20t)
everywhere inside it.
V = -d/dt (BA) = -d/dt[(4E-7 Tm/A)(500 turns/m)(4sin(20t) A)(.025m)2]
= [(4E-7 Tm/A)(500 turns/m)(80cos(20t) A/s)(.025m)2]
= 9.87E-5 cos(20t) Tm2/s = 9.87E-5 cos(20t)V
(d) For a current of 2 Amps in solenoid A, calculate the magnetic field inside solenoid A.
B = onAI = 4E-7 Tm/A)(1000 turns/m)(2A) = 2.51E-3 T
(e) A proton from a cosmic ray enters the field on a path perpendicular to the axis of the
solenoid. Calculate the radius of the path of the proton in the field.
p = Ber; r = mv/Be = (1.67E-27 kg )v/[(2.51E-3 T)(1.6E-19 C)]
Oops!, use v = 2.4 E5 m/s
r= (1.67)(2.4)/[(2.51)(1.6)] kg m/(s T C)
= 1.00 m
[B] Electrons are held in a circular path of 2 km. and are moving at a velocity of
3.0E6 m/s. A total of 10E10 electrons are in the beam.
(a) This beam has what current?
Current = A = C/s, Here, this is given by
Nqv/2πr = (1.6E-19 C)(10E10)(3.0E6 m/s)/(2π 2E3 m) = 3.8E-7 A
(b) What magnetic field does this beam produce at the center of the 2km ring?
B field at center of a ring with current I: B = μoI/2R
B = (4πE-7 Tm/A)(3.8E-7 A)/(4E3 m) = 1.19E-16 T
(c) What is the cyclotron frequency for the electrons in this beam?
2π v/2π r =  = V/r
 = (3.0E6 m/s)/(2E3m) = 1.50E3 rad/s
(d) To keep the electron beam in its circular path, what magnetic field must be used and
what direction does it point if the beam circulates clockwise?
Clockwise beam of electrons = CCW current. Right hand rule for IxB=F with
F toward center of circle gives B downward.
Now use p=Ber, B = mv/(er) = (9.11E-31 kg)(3.0E-6 m/s)/[(1.6E-19 C)(2E3 m)]
= 8.54E-21 kg/Cs = 8.54E-21 T
(e) If a magnetic field of 0.2 Tesla is produced with a solenoid carrying a current of 5
Amps, what can you say about the construction of the solenoid (number of coils, length,
area,..)?
For a solenoid, B=μo n I where n = turns/m.
So we have 0.2T = (4πE-7 T m/A) n (5 A)
Gathering numbers to the left,
3.18E4 /m = n = turns per meter.
Do one of questions C and D, 20 points.
[C] A square loop, a cm x b cm, lies on a table with 2 wires. One wire carries 2 amps, is
parallel to the side of length b and is y cm away from the closest side. The second wire
carries 5 amps, is parallel to the side of length a and is x cm away.
(a) Draw a diagram.
(b) Derive the expression for the magnetic field produced by each current.
B = oI/(2r) so, for top, B1 = o2/(2r1) = o/(r1) into page
for side, B2 = o5/(2r2) out of page.
(c) Derive the total magnetic flux in the loop using superposition.
For top wire, B1 the same for horizontal strips in rectangle,
so, B = o/(r1), dA = bdr1 and limits are r1 = y to y+a
1 = y y+a o/(r1) bdr1 = ob/ y y+a dr1 / r1
= (ob/) ln([y+a]/a) into page
For side wire, B2 the same for vertical strips in rectangle,
so, B = 5o/(2r2), dA = adr2 and limits are r2 = x to x+b
2 = x x+b 5o/(2r2) adr2 = 5oa/2 x x+b dr2 / r2
= (5oa/2) ln([x+b]/b) out of page
Superposition:
Add, but in opposite direction. Call out of page positive,
 = 2 - 1 = (5oa/2) ln([x+b]/b) - (ob/) ln([y+a]/a) out of page.
[D] A long solenoid with n turns per meter carries current I in the wire. The solenoid has
radius R.
(a) Use Ampere’s Law to find the magnetic field inside the solenoid.
∫Bds = μo I enclosed
B is zero outside the solenoid and
perpendicular to the integration path
for the vertical legs.
Inside the solenoid, B and ds are
parallel, giving ∫Bds = B L = μo I
where L = length of leg inside solenoid. For I, there are N turns inside the
integration path, each carrying current I, so we have
BL = μo NI
B = μo NI/L = μo nI n = N/L
(b) Determine the magnetic flux inside the solenoid.
Since the above result is independent of where in the solenoid the lower leg
of the loop is, B is uniform and perpendicular to the cross section (parallel to
dA) so we have
 = ∫BdA = B∫dA = BA = μo nIπR2
(c) A particle with mass m and charge q enters the field along a path making an angle of
45 with the axis of the solenoid. Find an expression for its trajectory. You can split the
trajectory into two motions to make it simpler.
Split the velocity into components parallel to and perpendicular to the
magnetic field direction. (In this case, both = v/√2). Call the direction of the
B field z.
The parallel component remains unchanged, so z = v/√2 t.
For the other component, p=Ber applies; here,
mv/√2 = Bqr
r = mv/(√2 Bq) gives the radius of the path and
2πR /(v/√2 ) gives the period = 2π√2 R/v