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
j.
particles acceleration produced by a uniform magneti
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 A 100 A
g = 3.3 10
-6
T = 3.3 T.
(b) This is about one
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 sides must be negative and exactly cancel the total of
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) Find the emf of
the battery. (b) Is the polarity shown cor
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 magnetic eld B
has a magnitude of B = 0.591 T and is horizon
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 at point
P due to two 1.50-mm segments of wire that ar
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 the cylinder?
(b) In terms of , what is the magnitude of
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 outer conductor with a radius of
3.7 mm. The two condu
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 the positive z axis, we can write the total B at
P as
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 A, calculate the magnetic eld in the solenoid. (b) Calc
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 with a magnetic
material. The magnetic eld inside the w
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, what are the magnitude and direction of the eld B at
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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 LED
Christmas tree lights were used for comparison. Th
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.
_
Abstract
The purpose of this experiment was to examin
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 on this assignment.
Abstract
The purpose of this experi
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 = 134 10-16 J. . . 2 2
c
hc
h
This is (1.34 10 16 J) / (1.6
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 electrons N is given by q = Ne, where e is the magnitude of
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 shown. The right-hand side of the circuit extends i
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 relative to the muon. Specifically, t0 = 2.2000 s. We a
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. The wavelength is related to the frequency by f = c,
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) The ground-state energy is
( 6.63 10 J s ) h2 E1 = n2
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 magnetic quantum number m ranges from - to + . For = 1 , t
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 is also the number of conduction electrons per unit volu
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 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K 5.30 106
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 note that the sum of superscripts (mass numbers A) must