Magnetic field and Induced Electric field inside a solenoid

Click For Summary
SUMMARY

The discussion focuses on calculating the magnetic field and induced electric field inside a solenoid with specific parameters: N=500 windings, radius a=0.1 m, height h=0.6 m, and a time-varying current I(t) = Io - bt, where Io = 0.4 amps and b = 0.2 amps/second. The magnetic field B(t) is derived using Ampere's Law, resulting in B(t) = (μ * N / L)(Io - bt). The electric field E is calculated using Faraday's Law, leading to E = (r/2) * (μo * N / L) * dI/dt, where dI/dt is determined as -b. The calculations and derivations presented are confirmed as correct by participants in the discussion.

PREREQUISITES
  • Understanding of Ampere's Law and its application in electromagnetism
  • Familiarity with Faraday's Law of electromagnetic induction
  • Basic knowledge of calculus, specifically differentiation
  • Concept of electromagnetic fields in solenoids
NEXT STEPS
  • Study the derivation of electromagnetic fields in solenoids using Ampere's Law
  • Learn about the applications of Faraday's Law in real-world scenarios
  • Explore the relationship between current changes and induced electric fields
  • Investigate the concept of electromagnetic energy and its calculations
USEFUL FOR

Physics students, electrical engineers, and anyone interested in the principles of electromagnetism and solenoid applications.

m0t0xk1d
Messages
1
Reaction score
0

Homework Statement


The problem give is: A solenoid has N=500 windings, radius a=.1 m and a height h = .6m; the current is found to be decreasing according to I(t) = Io - bt, where Io = .4 amps and b = .2 amps/second.
Calculate the rate at which electromagnetic energy is leaving the solenoid at t=1 second. Answer this overarching question by answering the following set of guided questions.

1. using amperes law derive an expression for the magnetic field inside the solenoid. your expression for the magnetic field with be a function of time.
2. use Faraday's law to calculate the electric field inside the solenoid

Homework Equations


Faradays law = ∫E*dl = -d\Phi / dt
amperes law = ∫B*dl = \muo * I

The Attempt at a Solution


my questions are for part 1 and 2 not the actually over hanging question of the energy.

For part 1 my attempt was amperes law = ∫B*dl = \muo * I. so B*L = \mu * I. Then B = ( \mu * I *N)/L. So once I got to here, to make B a function of time (B(t)) I plugged in the equation for I(t) for I and got B(t) = ( \mu * N / L ) (Io - bt). I have no way to see if this is right or wrong so I wanted to see if someone could check my work.


So for part 2 I got ∫E*dl = -d\Phi / dt = ∫E*dl = A dB/dt. then E *2\pir = A dB/dt. Then doing some algebra I got E = R/2 *db/dt = r/2 d/dt(\muoNI/L) from there on I went on to E = r/2 * \muo N/L * dI/dt. This is where it really stumps me. I don't know what to do with dI/dt. my original equation for I = Io - bt. My first thought was take the derivative which I came dI/dt = 1 - bt, but then I came to thinking b is .2 amps/second, which is already the rate at which I is decreasing, So I thought dI/dt = -b. but either was I was still unsure and looking to see if someone could help me out on this
 
Physics news on Phys.org
Part 1 looks correct to me.

Part 2, the formula for E looks correct but dI/dt is the rate of change of I with respect to time. So you have to take the first derivative of I. I think looking at the units as you started to do will confirm the answer for you.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
1K
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K