Clarification on induced current/Bfields for solenoids

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In summary: When dividing the magnetic flux by 100, you are accounting for the number of turns in the coil. However, in problem 2, you are dividing the flux by the area of the coil, not the number of turns. This is why you are off by a factor of 100.
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aimslin22
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Homework Statement


I'm working on induced currents/b-fields from Lenz's Laws and two different problems use the same equation but one has an extra variable and I don't know when to add this equation.

Problem 1:
In a physics laboratory experiment, a coil with 180 turns enclosing an area of 11.6cm^2 is rotated during the time interval 3.10×10−2s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00×10−5T. What is the total magnitude of the magnetic flux ( Phi_initial) through the coil before it is rotated?

Problem 1 solution equation:
Phi = BAcostheta
This relation is just for one loop, and when we are calculating the effect of all the loops in the coil we must remember to multiply by N, the total number of loops. In other words Phi_initial = N*Phi_1.

^Now this completely makes sense to me, but I just did another problem:

Problem 2:
A 100-turn 8 cm diameter coil is made of 0.5 mm diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B increase to induce a 2 A current in the coil?

I did this problem perfectly, but I I'm off by a factor of 100 because I divide my flux by 100 to account for the numbers of turns.

My question is: why do I need to account for the turns in one problem and not another? ie. In the future, how will I know what to do?

Thank you!
 
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  • #2
When you divide the flux that you have obtained by 100, you are finding the flux liked with each turn. But the net emf induced will be because of the net change in flux. i.e. the sum of flux change linked with each of the 100 coils.
 
  • #3
aimslin22 said:
1. Problem 1 solution equation:
Phi = BAcostheta
This relation is just for one loop, and when we are calculating the effect of all the loops in the coil we must remember to multiply by N, the total number of loops. In other words Phi_initial = N*Phi_1.
.
No. the flux phi is stll BA. It's only the emf that will get multiplied by 100: emf = -N d(phi)/dt.
Problem 2:
A 100-turn 8 cm diameter coil is made of 0.5 mm diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B increase to induce a 2 A current in the coil?

I did this problem perfectly, but I I'm off by a factor of 100 because I divide my flux by 100 to account for the numbers of turns.
We have
emf = -N d(phi)/dt and then i = emf/R so you need to know the resistance of the coil also. You can compute R from the wire length, diameter and conductivity of Cu.

Phi = B*A, A = area, so emf = -NA dB/dt. Solve for dB/dt.

The solutions of problems 1 and 2 are perfectly consistent.
 

FAQ: Clarification on induced current/Bfields for solenoids

What is induced current in a solenoid?

Induced current in a solenoid refers to the flow of electric charge that is created when a changing magnetic field passes through the coil of the solenoid. This current is induced because of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an electric current in a nearby conductor.

What is the direction of induced current in a solenoid?

The direction of induced current in a solenoid is determined by Lenz's law, which states that the induced current will flow in a direction that opposes the change in magnetic field that caused it. In a solenoid, this means that the induced current will flow in the opposite direction of the changing magnetic field passing through the coil.

How does the number of turns in a solenoid affect induced current?

The number of turns in a solenoid affects the strength of the induced current. More turns in the coil will create a stronger magnetic field, which in turn will induce a stronger current. This relationship between the number of turns and the strength of the current is known as the "turns ratio."

What is the relationship between magnetic field strength and induced current in a solenoid?

The strength of the induced current in a solenoid is directly proportional to the strength of the magnetic field passing through the coil of the solenoid. This means that a stronger magnetic field will induce a stronger current, while a weaker magnetic field will induce a weaker current.

How does the rate of change of magnetic field affect induced current in a solenoid?

The rate of change of magnetic field has a direct effect on the strength of the induced current in a solenoid. A faster rate of change will induce a stronger current, while a slower rate of change will induce a weaker current. This is because a faster rate of change will result in a stronger magnetic field passing through the coil, which will in turn induce a stronger current.

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