Rotating Square Loop in Constant B-field

AI Thread Summary
A square loop with a side length of 20 cm rotates in a magnetic field of 2.0 T, and the angle between the field and the loop's normal is 20° at a rate of 10°/s. The induced emf is calculated using the formula ε = -dΦ/dt, where Φ is the magnetic flux. The user consistently arrives at an answer of 0.27 V but questions its correctness. A clarification indicates that the angular velocity must be converted to radians per second for accurate calculations. The discussion concludes with a reminder about proper unit conversion in physics problems.
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[SOLVED] Rotating Square Loop in Constant B-field

Homework Statement


A square loop (length along one side = 20 cm) rotates in a constant magnetic field which has a magnitude of 2.0 T. At an instant when the angle between the field and the normal to the plane of the loop is equal to 20° and increasing at the rate of 10°/s, what is the magnitude of the induced emf in the loop?

a. 13mV
b. 0.27V
c. 4.8mV
d. 14mV
e. 2.2mV

Homework Equations



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

\Phi = BAcos(\theta) = BAcos(\omegat)

d\Phi = -BA\omegasin(\omegat)

The Attempt at a Solution



I'm trying to study for an exam and I've got this practice question that I can answer but my answer never matches. I keep getting b as an answer and I'm not sure if it's right.

\epsilon = BAcos(\omegat) = (2T)(0.2m)2(10)sin(10t)

I use t = 2s since it asks for \theta = 20° and I get

\epsilon = (2T)(0.2m)2(10)sin(20) ≈ 0.27 V

Am I making a mistake or a wrong assumption anywhere or could it be the answer is incorrectly marked?

Thanks!
 
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Using "10" might be fine inside the argument of the sine function, but if you're taking it outside when you take the derivative, you need to convert it to rad/s.
 
Thanks, didn't think of that.
 
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