Induced emf: Faraday's Law and Lenz's Law Question

AI Thread Summary
The discussion focuses on calculating the average induced electromotive force (emf) in a conductive loop as the magnetic field drops to zero. The participants confirm that the induced emf is calculated using Faraday's Law, resulting in a value of 1.33 V. They derive the formula for emf as ε = ΔΦ/Δt, leading to the same conclusion. Lenz's Law is briefly mentioned, indicating some confusion about its application in this context. Overall, the calculations and understanding of the principles involved are affirmed by multiple participants.
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[SOLVED] Induced emf: Faraday's Law and Lenz's Law Question

The plane of a conductive loop with an area of 0.020 m^2 is perpendicular to a uniform magnetic field of 0.30 T. If the field drops to zero in 0.0045 s, what is the magnitude of the average emf induced in the loop?






Given that,
Area (a) = 0.020m2
Magnetic filed (B) = 0.30 T.
When the field drops to zero,
Time taken = 0.0045 s
I have,
Induced emf = Change in flux
Time taken
= B x a cos 90
Time taken
= 0.30 x 0.020 x 1
0.0045
= 1.33 V


Is my work/answer correct?
 
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That should be correct.
 
rock.freak667 said:
That should be correct.

Is it really correct...or are you just telling me that?
 
I calculated it the same way as you and got the same answer

E=\frac{BA}{t}
 
Ok...thanks. I just can't believe that I did it right. :)
 
\Phi = \iint \mathbf{B} \cdot d\mathbf{a}

which in a constant magnetic field perpindicular to a constant area will reduce to

\Phi = BA

Then you know that

\epsilon = \frac{d \Phi}{dt}

So the emf would be (assuming a constant change in the magnetic field)

\epsilon = \frac{\Delta \Phi}{\Delta t}

and further shows that

\epsilon = \frac{A * (B_i - B_f)}{(t_i - t_f)}

where B_f and t_f are zero.

With numbers

\epsilon = \frac{.02*.3}{.0045}\frac{mB}{s^2} = 1.33V

You need some more confidence in yourself.

Edit:
Oh right, what about Lenz's law?
 
Last edited:
When it comes to physics I get so confused, that's why I'm on here. lol But thanks for your help.
 
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