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# Magnetic flux question?

A 200 turn coil has a 15.2 V potential difference induced in it when the magnetic field changes from 0.42 T to 0.22 T in the opposite direction in 3.2 x 10^2 s. What is the radius of the coil?

so i did:

E=-N delta phi/delta time

and came out with 121.6 for A.

then i did:

121.6=pie r^2

and got 6.22 but its wrong.

the real answer is 3.5 x10^2.

any help on this is much appreciated.

### 2 Answers

- JacyLv 71 decade agoFavorite Answer
Employ Faradays’ law of induction:

emf = dΦ/dt,

where: Φ = NBA

therefore: emf = NAdB/dt,

where: dB/dt = [0.42 – (-0.22)]/(3.2x10-²)

= 0.64/3.2 x 10-² = 0.2 x 10² T/s

Hence:

Emf = 15.2 = 200(πr²)(0.2x10²),

Therefore:

r = √ [15.2x10-²/(40π) = 3.5x10-² m

Note: I had to make a correction in the time the magnetic

field changes, I believe it should be changing in

a more realistic 3.2x10-² s and not 3.2x10² sec. as

you specified in your question.

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