Clarification on induced current/Bfields for solenoids

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SUMMARY

This discussion clarifies the application of Lenz's Law in calculating induced currents and magnetic fields in solenoids. In Problem 1, the magnetic flux through a coil with 180 turns and an area of 11.6 cm² is calculated using the equation Φ = BAcostheta, where Φ_initial is determined by multiplying the flux of a single loop by the number of turns (N). In Problem 2, a 100-turn coil with an 8 cm diameter requires understanding the induced electromotive force (emf) and the resistance of the coil to determine the rate of change of the magnetic field (dB/dt) necessary to induce a 2 A current. The key takeaway is that the total emf is influenced by the number of turns, while the magnetic flux remains constant for each loop.

PREREQUISITES
  • Understanding of Lenz's Law and electromagnetic induction
  • Familiarity with magnetic flux calculations (Φ = BAcostheta)
  • Knowledge of coil resistance calculations based on wire properties
  • Basic principles of electromotive force (emf) and current
NEXT STEPS
  • Study the principles of electromagnetic induction in depth
  • Learn how to calculate the resistance of a coil using wire diameter and material conductivity
  • Explore the relationship between induced emf and the rate of change of magnetic flux
  • Investigate practical applications of solenoids in electrical engineering
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Physics students, electrical engineers, and educators seeking to deepen their understanding of electromagnetic induction and its applications in solenoids and coils.

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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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.
 
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.
 

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