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3.3.3 In Class or Homework Exercise
1. The rectangular loop below is being pulled to the right, out of the magnetic field which points inward as shown. In what direction is the induced current?

No current is induced in the top or bottom wire, since they are moving parallel to themselves. The only current induced is in the left wire, where the current is induced upward. The result will be a clockwise current.

2. If the solenoid below is being pulled away from the loop shown, in what direction is the induced current in the part of the loop closest to the viewer?

First of all, use the second right hand rule to find the north pole of the coil. Doing this we determine that the north pole is the left end. Since we are pulling the north pole away from the loop, the right side of the loop must become a south pole (in order to try and pull the coil back). This means that current must have been induced in a direction up the front of the loop. 3. The magnetic flux through a coil of wire containing 2 loops changes from -20 Wb to +15 Wb in 1.4 s. What is the induced emf ?

i  20Wb

 f  15Wb
N 2
t  1.4 s
V ?

t
15  (20)
 (2)
1.4

V  N

 50V

UNIT 3 Fields

RRHS PHYSICS

Page 141 of 160

4. A square coil of sides 5.0 cm contains 10 0 loops and is positioned perpendicular to a uniform 0.60 T magnetic field. It is quickly and uniformly pulled from the field (moving perpendicularly to B) to a region where B drops abruptly to zero. It takes 0.10 s for the whole coil to reach the field fr ee region. How much energy is dissipated in the coil if its resistance is 100.0  ? How much work was done in pulling the coil out of the field?

A  (0.05) 2  0.0025m 2
N  100
B  0.60T
t  0.10 s
R  100.0
E  ?
W ?
   f  i
 0  BA
 (0.60)(0.0025)
 0.0015Wb

V  N


t

 (100)

0.0015
0.10

 1.5V

  Pt
 IVt
V
 Vt
R
V2

t
R
(1.5) 2

)(0.10)
100.0
 2.25  103 J
The amount of work...
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