The diagram on the right shows a magnetic field that always points in the direct
ID: 1322784 • Letter: T
Question
The diagram on the right shows a magnetic field that always points in the direction indicated hut has a magnitude that varies according to the graph on the left, A coil of wire with 25 turns (only one is shown) and an overall diameter of4.0 cm is situated in the magnetic field. Using the graph (which plots field strength. not flux), find the magnitude of the induced emf in the coil at a) t = 15 ms and Otheta =0 (i.e. the plane of the coil is perpendicular to the magnetic field), and at h r = 25 ms and 0 0. For c), use Theta = 50.Degree and assume the wire has a resistance of 0.27 Omega. Find the magnitude and direction (clockwise or counterclockwise from Snoopy?s point of view) of the induced current in the coil at time t = 25 ins. Hint: For part e) you will need Ohm?s Law: assume any resistance in the portion? of the circuit outside the coil is negligible.Explanation / Answer
Note that
emf = NA (dB/dt) cos (theta)
where
N = number of turns (25)
A = cross section area of the coil
dB/dt = rate change in B, which is the slope of the B-t graph
theta = the angle between the cross section and the B field
To get A,
A = pi r^2, where r = 4.0 cm/2 = 2.0 cm = 0.02 m
A = 0.001257 m^2
Plugging in all our data,
PART A:
For t = 15 m/s, theta = 0:
From the graph,
dB/dt = 0 [horizontal line]
Thus,
emf = 0 V [ANSWER]
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PART B:
For t = 25 ms,
dB/dt = slope = delta B/ delta T = 0.02 T/0.01 s
= 2 T/s
Thus, as theta = 0, cos(theta) = 1:
emf = NA (dB/dt) cos (theta)
emf = 0.0628 V or 62.8 mV [ANSWER]
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PART C:
For theta = 50 degrees, still at 25 ms,
emf = 0.0404 V or 40.4 mV
Thus, as V = IR ---> I = V/R, R = 0.27 ohms,
I = 0.15 A (clockwise relative to Snoopy) [ANSWER]
DONE! It was nice working with you!