What is the current induced in a wire loop placed in a changing magnetic field?

[moenste]

Homework Statement

A closed wire loop in the form of a square of side 4 cm is placed with its plane perpendicular to a uniform magnetic field, which is increasing at the rate of 0.3 T s^-1. The loop has negligible inductance, and a resistance of 2 * 10^-3 Ω. Calculate the current induced in the loop, and explain with the aid of a clear diagram the relation between the diraction of the induced current and the direction of the magnetic field.

Answer: 0.24 A.

2. The attempt at a solution
Well, I used the E = B L v and I = V / R formulas.

E = 0.3 * (4 / 100) = 0.012 V.

I = 0.012 / (2 * 10^-3) = 6 A.

What's wrong here?

 

[LemmeThink]

Hi!

Here's a hint -

E = B L v

Does this formula seem to apply here? Could you please list the value of each of these variables?

 

[moenste]

Hi!

Here's a hint -

Does this formula seem to apply here? Could you please list the value of each of these variables?

Hi!

E = voltage, B = magnetic field, L = length, v = velocity.

If know that this formula is not the best fit for this problem (we don't know velocity). But which one do you suggest?

 

[LemmeThink]

Do you know Faraday's law of induction?

 

[moenste]

Do you know Faraday's law of induction?

Got it from here. Thank you!

E = - N (d Φ / d t)

d Φ / d t = 0.3 T s^-1 * A, where A = (4 / 100) * (4 / 100) = 1.6 * 10^-3 m².

E= - 1 * 0.3 * 1.6 * 10^-3 = - 4.8 * 10^-4 V.

I = V / R = 4.8 * 10^-4 / 2 * 10^-3 = 0.24 A.