long solenoid (r=3cm, 2500 turns per meter) carries a current I=.3sin(200t)Amps,
ID: 2017348 • Letter: L
Question
long solenoid (r=3cm, 2500 turns per meter) carries a current I=.3sin(200t)Amps, where t is in seconds. When t= 5ms, what is the magnitude of the induced electric field at a point 2cm from the axis of the solenoid?
this is the answer
Given:
radius of the solenoid r = 0.03 m
E = -d(N)/dt = -L dI/dt ......... (1)
given dI/dt = 0.160 A/s What is givin about .160A/s where did this come from ?
L = (N^2A)/l ....... (2)
0 = 4 x 10-7 T m/A
N = 2500 x 0.0046 =11.5 where did .0046 come from
11.5A = pi (0.026)^2 = 0.002826 m2
l = 0.02 m
plug the values in equation 2 we have
L = (4 x 10-7 T m/A)(11.5)2(0.002826 m2)/(0.02)
= 2.34*10-5
now from the equation 1 induced emf not sure what happens after here ???
E = - L (di/dt)
= -2.34*10-5(0.016)
E = 3.8 x 10-3 volts
Explanation / Answer
I think there might be a few misprints in your question. To answer the first highlighted question about 0.16 A/s, I think this is supposed to read 0.016 A/s. Notice how they use this revised number at the bottom when they figure out the induced emf? If you plug in 0.005 (that's 5 ms, which is given in the problem) for "t" in the equation I = 0.3sin(200t), then you would get 0.016. This is what they are using for di/dt at the bottom.
The second highlighted question: 0.0046 is refering to the length of the solenoid. Since your question starts with a partial sentence, I think the text telling you how long it is (4.6 mm) got cut off?
The third question: I know this is going to sound crazy, but the answer is off, too. I've worked this problem over multiple times, and I keep getting 3.8e-7 volts.
Hope this helps. Please send a comment or something once you really do solve this mystery! I'd be interested to know if it truly was typos!
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