B t . The necessary information is given, except for B, the change in the magnet
ID: 1496080 • Letter: B
Question
B
t
.
The necessary information is given, except for B, the change in the magnetic flux during the elapsed time. Using the normal direction to coincide with the positive z-axis, compute the initial and final magnetic fluxes with
B BA = BA cos ,
find the difference, and assemble all terms in Faraday's law. The current can then be found with Ohm's law, and its direction with Lenz's law.
SOLUTION
(a) Find the induced emf in the coil.
To compute the flux, the area of the coil is needed.
A = L2 = (0.0180 m)2 = 3.24 10-4 m2
The magnetic flux
B,i
through the coil at t = 0 is zero because B = 0. Calculate the flux at t = 0.800 s.
B,f = BA cos = (0.500 T)(3.24 10-4 m2)cos(0°)
= 1.62 10-4 Wb
Compute the change in the magnetic flux through the cross section of the coil over the 0.800-s interval.
B = B,f B,i = 1.62 10-4 Wb
Substitute into Faraday's law of induction to find the induced emf in the coil.
[e m f] = N
B
t
= (25 turns)
1.62 10-4 Wb
0.800 s
= 5.06 10-3 V
(b) Find the magnitude of the induced current in the coil.
Substitute the voltage difference and the resistance into Ohm's law, where
V = [e m f] .
I =
V
R
=
5.06 10-3 V
0.350
= 1.45 10-2 A
(c) Find the direction of the induced current in the coil.
The magnetic field is increasing up through the loop, in the same direction as the normal to the plane; hence, the flux is positive and is also increasing. A downward-pointing induced magnetic field will create negative flux, opposing the change. If you point your right thumb in the clockwise direction along the loop as viewed from above, your fingers curl down through the loop, which is the correct direction for the counter magnetic field. Hence the current must proceed in a clockwise direction as viewed from above the coil.
PRACTICE IT
Use the worked example above to help you solve this problem. A coil with 37 turns of wire is wrapped on a frame with a square cross-section 2.16 cm on a side. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 0.542 . An applied uniform magnetic field is perpendicular to the plane of the coil, as in the figure.
(a) If the field changes uniformly from 0.00 T to 0.415 T in 0.865 s, find the induced emf in the coil while the field is changing.
[e m f] = -8.276 * 10^-3 [Correct: Your answer is correct.] V
(b) Find the magnitude of the induced current in the coil while the field is changing.
I = 1.526* 10^-2 [Correct: Your answer is correct.] A
EXERCISE
Suppose the magnetic field changes uniformly from 0.415 T to 0.290 T in the next 0.585 s.
(a) Compute the induced emf in the coil.
[e m f] = ____ V
(b) Compute the induced current.
I = ____ A counterclockwise as viewed from above the coil
Please help calculate the EXERCISE portion. Show work and answer.
Explanation / Answer
Hi,
Well, in the part EXERCISE you can procede the same way you did it during the previous part.
We use Faraday's Law to find the value of the emf and then we use Ohm's Law to calculate the value of the current with the resistance given.
We are going to assume that the data for this exercise are as follows:
R = 0.542 ; N = 37 ; a = 2.16 cm2 (A = 4.67*10-4 m2) ; t = 0.585 s ; B1 = 0.415 T ; B2 = 0.290 T
The magnetic flux in this case can be calculated as:
= BA ; where B is the magnetic field and A is the cross sectional area of the coil
1 = (0.415 T)*(4.67*10-4 m2) = 1.94*10-4 Wb
2 = (0.290 T )*(4.67*10-4 m2) = 1.35*10-4 Wb
= 2 - 1 = 1.35*10-4 Wb - 1.94*10-4 Wb = -5.86*10-5 Wb
(a) emf = -N /t = -37 *(-5.86*10-5 Wb/0.585 s) = 3.71*10-3 V
(b) I = (3.71*10-3 V)/(0.542 ) = 6.84*10-3 A
I hope it helps.
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