A problem dealing with Faraday's Low of Induction

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The discussion focuses on calculating the current in a coil using Faraday's Law of Induction. A 22.0 cm diameter coil with 20 turns of copper wire (2.6 mm diameter) experiences a changing magnetic field at a rate of 8.65 x 10^-3 T/s. The resistance is calculated using the formula R = (ρL)/A, resulting in R = 6.96 x 10^-4 ohms. The electromotive force (Emf) is determined to be 9.19 x 10^-7 volts, leading to a final current of 0.001 A.

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jena
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Hi,

My question:

A 22.0 cm diameter coil consists of 20 turns of circular cooper wire 2.6mm in diameter. a uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.65 x 10^-3 T/s. Determine the current in the loop.

I'm not sure how to approach this equation. I was thinking I should use the following equations:

R=((rho)L)/A, to find the resistance

Emf=-N(magnetic flux)/t, for the voltage

then using the resistance and voltage that I found from the equations use them to find the current, I.

I=(V)/R

Is anyone of this correct:confused:

Thank You:smile:
 
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I think there are some typos in your question or I don't understand how something can have 2 diameters at the same time. In addition, no resistivity was given.

A 22.0 cm diameter coil consists of 20 turns of circular cooper wire 2.6mm in diameter

But in general it looks like the right approach. Try working it out and see how the math goes.
 
Okay:smile:

Work:

Beginning with resistance
R=((rho)L)/A
R=((1.68 x 10^-8 ohm x m)(22 x10^-2 m))/(pi(2.6 x10^-3))^2
R= 6.96 x 10^-4 ohm

Next Emf for volts
E=-N(magnetic flux)/t,
E=-(20)(-8.65 x 10^-3 T/s)((pi(2.6 x10^-3))^2)
E= 9.19 x 10^-7 volts

Finally the current
I=E/R
I=(9.19 x 10^-7 volts)/(6.96 x 10^-4 ohm)
I=.001 A

Is this okay?

Thank You:smile:
 

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