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alexcoco
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[SOLVED] Rotating Square Loop in Constant B-field
[itex]\epsilon[/itex] = - [itex]\frac{d\Phi}{dt}[/itex]
[itex]\Phi[/itex] = BAcos([itex]\theta[/itex]) = BAcos([itex]\omega[/itex]t)
[itex]d\Phi[/itex] = -BA[itex]\omega[/itex]sin([itex]\omega[/itex]t)
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.
[itex]\epsilon[/itex] = BAcos([itex]\omega[/itex]t) = (2T)(0.2m)2(10)sin(10t)
I use t = 2s since it asks for [itex]\theta[/itex] = 20° and I get
[itex]\epsilon[/itex] = (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!
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
[itex]\epsilon[/itex] = - [itex]\frac{d\Phi}{dt}[/itex]
[itex]\Phi[/itex] = BAcos([itex]\theta[/itex]) = BAcos([itex]\omega[/itex]t)
[itex]d\Phi[/itex] = -BA[itex]\omega[/itex]sin([itex]\omega[/itex]t)
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.
[itex]\epsilon[/itex] = BAcos([itex]\omega[/itex]t) = (2T)(0.2m)2(10)sin(10t)
I use t = 2s since it asks for [itex]\theta[/itex] = 20° and I get
[itex]\epsilon[/itex] = (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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