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Problem 38 Hard Difficulty

A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Answer

a. $1.02 \times 10^{-4} \mathrm{V} / \mathrm{m}$
b. $2.04 \times 10^{-4} \mathrm{V} / \mathrm{m}$

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Video Transcript

we're told that we have a long, thin still annoyed with 900 terms per meters. That's in, of course, the primitive ity here mu not as 1.25 times 10 to the minus six Dupree appears, and we increase the sill. Annoyed at a uniform rate. D i d t then is 36 amperes percent. And the radius, which I call our is 0.2 for, uh, quiz 0 to 5 meters. Okay, then, to find, um, the, uh, electric field at a point near the center First, we're gonna use e. God d s is equal to minus the integral of G B DT That d s Okay, Well, the magnetic field b is equal to come You not times the number of turns times the current I therefore d b d t is equal to you not times the number of turns times d i d t we know all of those values. So then going back up to the integral the left side of the integral is the electric field times two pi r where the right side is equal to come you not times the number of turns times d I d T times pi r squared in general for the first case, if we're trying to solve for the center so we'll start another page here. So therefore the electric field is equal to you, not times the number of turns times D i d t multiplied by Are the pies cancel out, divided by two. So plucking all those values of the expression we find that near the center the electric field is equal to 5.9 times 10 to the minus. For and in the units here would be Newtons for Coolum so we can go ahead and box it in and then asked us to do two more things A and C. So for part A, it says Well, now let's consider it at a point. That's 20.0.5 meters from the center. So therefore he here is going to be equal to you, not times the number of turns times d i d t. But the integral here isn't from the center to the outside point are of the circle. The interval here is from the center to our minus 0.5 square. But of course, this is all gonna be divided by two times are so playing all those values into this expression, we find that this is equal to three 0.26 times 10 to the minus four news for cool Oh making box that in is their solution for part a. And then be It's the same thing now, but we're asked to consider a point that's 0.0.1 meters from the center. Okay, so then e is going to be equal to everything else is the same, except for 0.1 meters. Now, instead of 0.5 d i dt, multiplied by our minus 0.1 meters squared, divided by everything is divided by two plugging all those values into this expression, we find that the this value for the electric field comes out to equal 1.83 times, 10 to the minus four. Newton's her cool. Oh,

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