Rotating Square Loop in Constant B-field

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SUMMARY

The discussion centers on calculating the induced electromotive force (emf) in a rotating square loop within a constant magnetic field of 2.0 T. The loop has a side length of 20 cm and rotates such that the angle between the magnetic field and the loop's normal is 20°, increasing at a rate of 10°/s. The correct induced emf is calculated using the formula \(\epsilon = -\frac{d\Phi}{dt}\), leading to a result of approximately 0.27 V, confirming option b as the correct answer. The user initially miscalculated due to not converting degrees to radians for angular velocity.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Familiarity with the concept of magnetic flux
  • Knowledge of angular velocity and its conversion from degrees to radians
  • Basic proficiency in trigonometric functions, particularly sine
NEXT STEPS
  • Study the derivation and applications of Faraday's Law in electromagnetic systems
  • Learn about magnetic flux calculations in rotating systems
  • Explore the conversion of angular measurements from degrees to radians in physics
  • Investigate the effects of varying magnetic fields on induced emf
USEFUL FOR

Students preparing for physics exams, educators teaching electromagnetism, and anyone interested in the principles of electromagnetic induction and its applications in rotating systems.

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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



[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!
 
Last edited:
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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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