Induced Current problem from review

In summary, a 40 turn circular coil with a radius of 4.0 cm and a total resistance of 0.20 Ω is placed in a uniform magnetic field with a magnitude that varies over time according to B = 50*sin(10π*t) mT. Using the equations ε = -N*d(ΦB)/dt, I = V/R, and ΦB = ∫B*dA, the induced current in the coil at t=0.10 s is calculated as I = (-N*50*sin(10π*0.10)*10^-3*πr^2)/R, but this may not give the correct answer due to possible errors in setting up the magnetic field
  • #1
Boris_The_Red
1
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Homework Statement


A 40 turn circular coil of a 4.0 cm radius and a total resistance of 0.20 Ω is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as B = 50*sin(10π*t) mT where t is measured in s. What is the induced current in the coil at 0.10 s?

Homework Equations


I believe these are the equations needed:
ε = -N*d(ΦB)/dt
I = V/R
ΦB = ∫B*dA


The Attempt at a Solution


The way I did it was start off with I = V/R, then knowing that V is the emf, I plugged in ε = -N*d(ΦB)/dt.

Then I used the flux equation, but modified it since we were given dB (or what I assume is dB) and I thus get I = (-N* ∫dB*A)/ R.

From there I get I = (-N*50*sin(10π*t)*10-3 *πr2)/R

I solve this, but I don't get the correct answer (which is known since this is from an older exam that I'm doing as additional practice for my fnal).

Am I missing something key here, or did I improperly set up the mag. field equation?
 
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  • #2
Boris_The_Red said:
Then I used the flux equation, but modified it since we were given dB (or what I assume is dB) and I thus get I = (-N* ∫dB*A)/ R.
Why did you bring dB into the picture? The problem does not allude to dB in any way.
This problem could have been made really interesting if B = 0 for t < 0. But you don't want to consider that possibility. Just assume the B field has been there for a long time and then when the field is zero and starting to increase (B=0, dB/dt > 0) that defines t=0. Then you want the current 0.1sec. later.
 

1. What is an induced current problem?

An induced current problem is a scenario where a changing magnetic field induces an electric current in a conductor. This can happen when the conductor moves through a magnetic field, the magnetic field changes over time, or when a magnet is moved near a conductor.

2. How do you solve an induced current problem?

To solve an induced current problem, you need to use Faraday's law of induction, which states that the induced electromotive force (EMF) in a closed circuit is equal to the rate of change of the magnetic flux through the circuit. You can use this law to calculate the magnitude and direction of the induced current in the conductor.

3. What factors affect the magnitude of the induced current?

The magnitude of the induced current depends on several factors, including the strength of the magnetic field, the speed at which the conductor moves through the field, the angle between the magnetic field and the conductor, and the length and material of the conductor. These factors can affect the rate of change of the magnetic flux and therefore the magnitude of the induced current.

4. How does Lenz's law apply to induced current problems?

Lenz's law states that the direction of the induced current in a conductor will be such that it opposes the change in the magnetic field that caused it. This means that if the magnetic field is increasing, the induced current will flow in the opposite direction to the change in the field, and if the field is decreasing, the induced current will flow in the same direction as the change in the field.

5. Can induced current problems have practical applications?

Yes, induced current problems have many practical applications in our everyday lives. Some examples include generators that convert mechanical energy into electrical energy, transformers that change the voltage of an alternating current, and induction cooktops that use a changing magnetic field to generate heat in a conductor. Understanding induced currents is also crucial in the design of electric motors and generators used in various industries.

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