Induced EMF in N=181 Coil, A=.01955^2 at t=2.50, 7.50, 15.0, 25.0ms

[wr1015]

A magnetic field with the time dependence shown in Figure 23-38 is at right angles to a 181 turn circular coil with a diameter of 3.91 cm. What is the induced emf in the coil at each of the following times?

(a) t = 2.50 ms
0 V
(b) t = 7.50 ms

(c) t = 15.0 ms
0 V
(d) t = 25.0 ms

N = 181
A = .01955$$^2$$

i've gotten (a) and (c) right but am having a very hard time with (b) and (d). for (b) I've been doing $$\phi$$ = BA cos $$\theta$$ where $$\theta$$ = 0 for t= 7.50 m/s as the final flux and then at t= 2.50 m/s for the initial flux . Then to find induced emf I've been doing 181(($$\phi_{f}$$ - $$\phi_{i}$$)/(7.50 - 2.50)) but am getting the wrong answer, what am i doing wrong??

 

[wr1015]

can anyone help me?

 

[nrqed]

A magnetic field with the time dependence shown in Figure 23-38 is at right angles to a 181 turn circular coil with a diameter of 3.91 cm. What is the induced emf in the coil at each of the following times?

(a) t = 2.50 ms
0 V
(b) t = 7.50 ms

(c) t = 15.0 ms
0 V
(d) t = 25.0 ms

N = 181
A = .01955$$^2$$

i've gotten (a) and (c) right but am having a very hard time with (b) and (d). for (b) I've been doing $$\phi$$ = BA cos $$\theta$$ where $$\theta$$ = 0 for t= 7.50 m/s as the final flux and then at t= 2.50 m/s for the initial flux . Then to find induced emf I've been doing 181(($$\phi_{f}$$ - $$\phi_{i}$$)/(7.50 - 2.50)) but am getting the wrong answer, what am i doing wrong??

To be honest, I am really not following what you did.

But the induced emf is (dropping all signs) N A cos (theta) dB/dt in your case (only the magnitude of B changes). And dB/dt is simply the slope of the graph. So for the emf at 7.50 ms, the slope of the graph is dB/dt = (-0.01 - 0.02)/(10 ms - 5 ms) . Dropping the sign, multiplying by N A should give you the answer.


Patrick

 

[wr1015]

To be honest, I am really not following what you did.

But the induced emf is (dropping all signs) N A cos (theta) dB/dt in your case (only the magnitude of B changes). And dB/dt is simply the slope of the graph. So for the emf at 7.50 ms, the slope of the graph is dB/dt = (-0.01 - 0.02)/(10 ms - 5 ms) . Dropping the sign, multiplying by N A should give you the answer.Patrick

(-.01-.02)/(10-5) = -.006

dropping the sign: (181) (.006) $$\pi$$ (.01955$$^2$$)

i get .001303 V but that's not right

edit: fixed

 

[nrqed]

(-.01-.02)/(10-5) = -.006

dropping the sign: (181) (.006) (.01955$$^2$$)

i get .001303 V but that's not right

You seem tohave forgotten to multiply r^2 by Pi !

 

[wr1015]

You seem tohave forgotten to multiply r^2 by Pi !

oops sorry that was supposed to have $$\pi$$ in there, i didn't forget it in my calculation

 

[nrqed]

(-.01-.02)/(10-5) = -.006

dropping the sign: (181) (.006) $$\pi$$ (.01955$$^2$$)

i get .001303 V but that's not right

edit: fixed

You must put dB/dt in Tesla per second...so it`s 6 Tesla/second

 

[wr1015]

You must put dB/dt in Tesla per second...so it`s 6 Tesla/second

ooooooohhhh ok, but why isn't it already in T/s??

 

[nrqed]

ooooooohhhh ok, but why isn't it already in T/s??

Because the time you divided by was in milliseconds.
Do you get the right answer now? I need to go to bed