Magnetic Fields from Currents
Josh Fuchs
Date of Experiment: 2/26/2014
Grishma Alakkat
Lab Partners: Molly Quillin, Chris Miller
I have neither given nor received unauthorized aid on this assignment.
Physics 21
Fall, 2012
Solution to HW-10
27-4 A particle with mass m = 1.81103 kg and a charge
of q = 1.22 108 C has, at a given instant, a velocity
v = (3.00 104 m/s) (a) What is the magnitude of the
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by
B=
0i
2r
.
With r = 20 ft = 6.10 m, we have
c4 10 B=
hb 2 b6.10 mg
-7
T m
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the
Physics 21
Fall, 2012
Solution to HW-9
26-32 In the circuit shown in the gure both batteries have
insignicant internal resistance and the idealized ammeter
reads 1.60 A in the direction shown. (a) Fin
Physics 21
Fall, 2012
Solution to HW-12
27-40 A straight, vertical wire carries a current of I =
1.19 A downward in a region between the poles of a large
superconducting electromagnet, where the magne
Physics 21
Fall, 2012
Solution to HW-13
28-14 Two parallel wires are 5.00 cm apart and carry currents in opposite directions, as shown in the gure. Find
the magnitude and direction of the magnetic eld
Physics 21
Fall, 2012
Solution to HW-14
22-31 An innitely long cylindrical conductor has radius
R and uniform surface charge density . (a) In terms of R
and , what is the charge per unit length for th
Physics 21
Fall, 2012
Solution to HW-7
evaluating the line integral of the electric eld. We will evaluate
f
Vf Vi =
24-12 A cylindrical capacitor has an inner conductor with
a radius of 1.8 mm and an
Physics 21
Fall, 2012
Solution to HW-15
28-37 Calculate the magnitude of the magnetic eld at
point P of the gure in terms of R, I1 , and I2 . What does
your expression give when I1 = I2 ?
direction of
Physics 21
Fall, 2012
Solution to HW-18
30-18 An air-lled toroidal solenoid has 320 turns of wire,
a mean radius of R = 13.0 cm, and a cross-sectional area of
A = 4.60 cm2 . (a) If the current is 5.9
Physics 21
Fall, 2012
Solution to HW-16
28-56 The current in the windings of a toroidal solenoid
is 2.400 A. There are N = 500 turns and the mean radius is
r = 25.00 cm. The toroidal solenoid is lled
Physics 21
Fall, 2012
Solution to HW-17
29-7 The current in the long, straight wire AB shown in
the gure is upward and is increasing steadily at a rate di/dt.
(a,b) At an instant when the current is i
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IMPORTANT: T
Grishma Alakkat
Abstract
The purpose of this experiment was to compare the brightness of light bulbs connected in
parallel and in series circuits. Incandescent light bulbs were used at first, and then
Electric Fields and Potentials
Josh Fuchs Section 504
Date of Experiment: January 23, 2014
Grishma Alakkat
Lab Partners: Walker OBrien, Mark Drosnes
I have neither given nor received unauthorized aid
Grishma Alakkat
PHYS 117
Josh Fuchs
February 25, 2014
Abstract
In this experiment, we measured the voltage of a capacitor at different times after it was
disconnected from the power source. We recorde
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .
c
hc
h
(b) The kinetic energy of the proton is
K=
2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 13
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of elec
Physics 117: HW- 7: Ch- 28; Due on 10- 6/7 before the class begins Problem 1. A wire of negligible resistance is bent into a rectangle as shown in the Figure with a battery and resisted connected as s
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 - 2 , and = v/c), we obtain
= 1-
FG t IJ . H t K
2 0
The proper time interval is measured by a clock at rest
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m. (b) It is in the infrared region.
2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfies
En L = En L
FG IJ = FG L IJ H K H L K
-2
2
=
1 , 2
which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.
2. (a
1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the mag
1. The number of atoms per unit volume is given by n = d / M , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n i
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:
rmin
-19 9 q qCu kq qCu ( 2e )( 29 ) (
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.
2. We