Download Mastering Physics Solutions to Week 10 11 Assignment

Physics 210 Mastering Physics Solutions Week 10 & 11 27.14.
 

IDENTIFY: When B is uniform across the surface,  B  B  A  BA cos  .

SET UP: A is normal to the surface and is directed outward from the enclosed volume. For


surface abcd, A   Aiˆ. For surface befc, A   Akˆ. For surface aefd, cos  3/5 and the flux is
positive.
 
EXECUTE: (a)  B (abcd )  B  A  0.
 
(b)  B (befc)  B  A  (0128 T)(0300 m)(0300 m)  00115 Wb.
 
(c)  B (aefd )  B  A  BA cos  53 (0128 T)(0500 m)(0300 m)  00115 Wb.
(d) The net flux through the rest of the surfaces is zero since they are parallel to the x-axis. The
total flux is the sum of all parts above, which is zero.
EVALUATE: The total flux through any closed surface, that encloses a volume, is zero.
27.10.
IDENTIFY: Knowing the area of a surface and the magnetic field it is in, we want to calculate
the flux through it.

 
SET UP: dA  dAkˆ , so d  B  B  dA  Bz dA.
EXECUTE:  B  Bz A  (0500 T)(00340 m) 2  578  104 T  m 2 .  B  5.78  104 Wb.
EVALUATE: Since the field is uniform over the surface, it is not necessary to integrate to find
the flux.
28.49.
(a) IDENTIFY and SET UP: The magnetic field near the center of a long solenoid is given by Eq.
(28.23), B  0 nI .
EXECUTE: Turns per unit length n 
B
0 I

00270 T
(4  107 T  m/A)(120 A)
 1790 turns/m
(b) N  nL  (1790 turns/m)(0400 m)  716 turns
Each turn of radius R has a length 2 R of wire. The total length of wire required is
N (2 R )  (716)(2 )(140  102 m)  630 m.
EVALUATE: A large length of wire is required. Due to the length of wire the solenoid will have
appreciable resistance.
29.11.
IDENTIFY: A change in magnetic flux through a coil induces an emf in the coil.
SET UP: The flux through a coil is  B  NBA cos and the induced emf is   d  B /dt.
EXECUTE: (a)   d  B /dt  d [ A( B0  bx)]/dt  bA dx /dt  bAv
(b) clockwise
(c) Same answers except the current is counterclockwise.
EVALUATE: Even though the coil remains within the magnetic field, the flux through it changes
because the strength of the field is changing.
29.27.
IDENTIFY and SET UP:   vBL. Use Lenz’s law to determine the direction of the induced
current. The force Fext required to maintain constant speed is equal and opposite to the force FI
that the magnetic field exerts on the rod because of the current in the rod.
EXECUTE: (a)   vBL  (7.50 m/s)(0.800 T)(0.500 m)  3.00 V

(b) B is into the page. The flux increases as the bar moves to the right, so the magnetic field of
the induced current is out of the page inside the circuit. To produce magnetic field in this
direction the induced current must be counterclockwise, so from b to a in the rod.
(c) I 

R


300 V
 200 A. FI  ILB sin   (200 A)(0500 m)(0800 T)sin 90  0800 N. FI is to the left.
150 
To keep the bar moving to the right at constant speed an external force with magnitude
Fext  0.800 N and directed to the right must be applied to the bar.
(d) The rate at which work is done by the force Fext is Fext v  (0800 N)(750 m/s)  600 W. The rate
at which thermal energy is developed in the circuit is I 2 R  (200 A)2 (150 )  600 W. These two
rates are equal, as is required by conservation of energy.
EVALUATE: The force on the rod due to the induced current is directed to oppose the motion of
the rod. This agrees with Lenz’s law.
29.64.
IDENTIFY: Apply
Faraday’s law to calculate the magnitude and direction of the induced emf.

SET UP: Let A be directed out of the page in Figure P29.64 in the textbook. This means that
counterclockwise emf is positive.
EXECUTE: (a) B  BA  B0 r02 (1  3(t /t0 ) 2  2(t /t0 )3 ).
(b)   
d B
d
B  r2
  B0 r02 (1  3(t /t0 ) 2  2(t /t0 )3 )   0 0 (6(t /t0 )  6(t /t0 ) 2 ).
dt
dt
t0
2
6B0 r02   t   t  
3
     . At t  5.0  10 s,
t0   t0   t0  


2

6 B  (0.0420 m) 2   5.0  103 s   5.0  103
  0

  
 0.010 s
0.010 s
  0.010 s

 

s 
  0.0665 V.  is positive so it is


counterclockwise.
(c) I 

Rtotal
 Rtotal  r  R 

I
r
0.0665 V
3.0  103 A
 12  10.2.
(d) Evaluating the emf at t  1.21 102 s and using the equations of part (b),   0.0676 V, and
the current flows clockwise, from b to a through the resistor.
  t 2  t  
  t0   t0  


(e)   0 when 0         . 1 
t
and t  t0  0.010 s.
t0

EVALUATE: At t  t0 , B  0. At t  5.00  103 s, B is in the  kˆ direction and is decreasing in

magnitude. Lenz’s law therefore says  is counterclockwise. At t  0.0121 s, B is in the  kˆ
direction
and is increasing in magnitude. Lenz’s law therefore says  is clockwise. These results for the
direction of  agree with the results we obtained from Faraday’s law.
29.72.
IDENTIFY: A bar moving in a magnetic field has an emf induced across its ends. The propeller
acts as such a bar.
SET UP: Different parts of the propeller are moving at different speeds, so we must integrate to
get the total induced emf. The potential induced across an element of length dx is d   vBdx, where
B is uniform.
EXECUTE: (a) Call x the distance from the center to an element of length dx, and L the length
of the propeller. The speed of dx is x, giving d   vBdx  x Bdx.   
L/2
0
x Bdx   BL2 /8.
(b) The potential difference is zero since the potential is the same at both ends of the propeller.
 220 rev 
4
(c)   (2 rad/rev) 
 (0.50  10 T)
 60 s 
(2.0 m) 2
 5.8  104 V  0.58 mV
8
EVALUATE: A potential difference of about
1
2
mV is not large enough to be concerned about in
a propeller.
29.7.IDENTIFY: Calculate the flux through the loop and apply Faraday’s law.
SET UP: To find the total flux integrate d  B over the width of the loop. The magnetic field of a
long straight wire, at distance r from the wire, is B 
right-hand rule.
EXECUTE: (a) B 

0 I
. The direction of B is given by the
2 r
0i
, into the page.
2 r
0i
Ldr.
2 r
b
 iL b dr 0iL
(c)  B   d  B  0 

ln(b/a ).
a
2 a r
2
d
 L
di
(d)   B  0 ln(b /a) .
2
dt
dt
 (0.240 m)
(e)   0
ln(0.360/0.120)(9.60 A/s)  5.06  107 V.
2
EVALUATE: The induced emf is proportional to the rate at which the current in the long straight
(b) d  B  BdA 
wire is changing
29.51.
IDENTIFY: The changing current in the solenoid will cause a changing magnetic field (and
hence changing flux) through the secondary winding, which will induce an emf in the secondary
coil.
SET UP: The magnetic field of the solenoid is B  0ni, and the induced emf is   N
dB
.
dt
EXECUTE: B  0ni  (4  107 T  m/A)(90.0  102 m 1)(0.160 A/s 2 )t 2  (1.810  103 T/s 2 )t 2 . The total
flux through secondary winding is (5.0) B (2.00  104 m 2 )  (1.810  106 Wb/s2 )t 2 .
 N
dB
 (3.619  106 V/s)t. i  3.20 A says 3.20 A  (0.160 A/s 2 )t 2 and t  4.472 s. This gives
dt
  (3.619  106 V/s)(4.472 s)  1.62  105 V.
EVALUATE: This a very small voltage, about 16  V.