MHB What is the current in a wire loop outside a solenoid with a changing current?

Click For Summary
SUMMARY

The discussion focuses on calculating the induced current in a wire loop outside a solenoid with a changing current. A 2000-turn solenoid, 2.0 m long and 15 cm in diameter, has a current increasing at 1.0 kA/s. For the 25-cm-diameter loop lying entirely outside the solenoid, the induced current was found to be 4.4 mA by considering only the magnetic field inside the solenoid, as the external magnetic field is negligible. The calculations utilize Faraday's law of electromagnetic induction, specifically the equation $\varepsilon = -\d{\Phi}{t}$.

PREREQUISITES
  • Understanding of Faraday's law of electromagnetic induction
  • Knowledge of solenoid properties and magnetic fields
  • Ability to perform calculations involving resistance and induced current
  • Familiarity with magnetic field concepts inside and outside solenoids
NEXT STEPS
  • Learn about the magnetic field inside a solenoid and its calculations
  • Study the application of Faraday's law in different geometries
  • Explore the concept of induced electromotive force (emf) in circuits
  • Investigate the effects of loop diameter on induced current in electromagnetic systems
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electromagnetic theory and applications, particularly in understanding induced currents in circuits involving solenoids.

MermaidWonders
Messages
112
Reaction score
0
A 2000-turn solenoid is 2.0 m long and 15 cm in diameter. The solenoid current is increasing at 1.0 kA/s.
(a) Find the current in the 10-cm-diameter wire loop with resistance 5.0 $\varOmega$ lying inside the solenoid and perpendicular to the solenoid axis.
(b) Repeat for a similarly-oriented 25-cm-diameter loop with the same resistance, lying entirely outside the solenoid.

I did part (a) fine, but I'm currently stuck on (b). There shouldn't be a very weak magnetic field outside the solenoid, right, so how can I find the values of the initial and final magnetic fields and the solenoid current outside it? If this isn't necessary, then how can I work around Faraday's law ($\varepsilon = -\d{\varPhi}{t}$) to get the new current of the wire loop?
 
Mathematics news on Phys.org
MermaidWonders said:
A 2000-turn solenoid is 2.0 m long and 15 cm in diameter. The solenoid current is increasing at 1.0 kA/s.
(a) Find the current in the 10-cm-diameter wire loop with resistance 5.0 $\varOmega$ lying inside the solenoid and perpendicular to the solenoid axis.
(b) Repeat for a similarly-oriented 25-cm-diameter loop with the same resistance, lying entirely outside the solenoid.

I did part (a) fine, but I'm currently stuck on (b). There shouldn't be a very weak magnetic field outside the solenoid, right, so how can I find the values of the initial and final magnetic fields and the solenoid current outside it? If this isn't necessary, then how can I work around Faraday's law ($\varepsilon = -\d{\varPhi}{t}$) to get the new current of the wire loop?

Suppose we neglect the contribution of the magnetic field outside the solenoid?
So we only consider the magnetic field inside the solenoid to find the induction in the loop outside the solenoid.
 
I like Serena said:
Suppose we neglect the contribution of the magnetic field outside the solenoid?
So we only consider the magnetic field inside the solenoid to find the induction in the loop outside the solenoid.

Sorry, but I'm totally confused here. :( So how come we consider the magnetic field inside the solenoid when the loop lies entirely outside it? Can you please explain?
 
MermaidWonders said:
Sorry, but I'm totally confused here. :( So how come we consider the magnetic field inside the solenoid when the loop lies entirely outside it? Can you please explain?

We have $\varepsilon = -\d \Phi t$. This applies to all of the magnetic field that passes through the loop. And that includes the magnetic field that is inside the solenoid, while the part outside the solenoid is presumably negligible.
 
I like Serena said:
We have $\varepsilon = -\d \Phi t$. This applies to all of the magnetic field that passes through the loop. And that includes the magnetic field that is inside the solenoid, while the part outside the solenoid is presumably negligible.

Wait, but the loop lies outside the solenoid... So sorry, but I'm still confused here. :(
 
MermaidWonders said:
Wait, but the loop lies outside the solenoid... So sorry, but I'm still confused here. :(

How about not letting ourselves be stopped by that, but just do the math?
 
I don't know how to go about doing this calculation, but when I did the math assuming that everything else is the same, I got 12 mA as the induced current, which is the wrong answer (I got that by just changing the radius of the loop). :(
 
OK... I just recalculated this, but I used the radius of the solenoid (15 cm / 2 = 7.5 cm = 0.075 m) this time around, since this 25-cm-diameter loop has a greater diameter than the solenoid, and since the magnetic field only applies for inside the solenoid. I got 4.4 mA this time, which is correct. My reasoning probably makes no sense whatsoever, but can you please explain why?
 
MermaidWonders said:
OK... I just recalculated this, but I used the radius of the solenoid (15 cm / 2 = 7.5 cm = 0.075 m) this time around, since this 25-cm-diameter loop has a greater diameter than the solenoid, and since the magnetic field only applies for inside the solenoid. I got 4.4 mA this time, which is correct. My reasoning probably makes no sense whatsoever, but can you please explain why?

Good! (Happy)

We have:
$$\epsilon = -\d \Phi t = -\d {(B_i A_i + B_o A_o)} t\approx -A_i\d {B_i} t $$
where $A_i$ is the cross sectional area inside the solenoid, $B_i$ is the magnetic field inside the solenoid, $A_o$ is the remaining area outside the solenoid, and $B_o\approx 0$ is the magnetic field outside the solenoid and inside the loop.

This is what we have in a loop outside and around the solenoid, where we neglect the reverse magnetic field, which we can as long as the loop is small enough.
 
  • #10
I like Serena said:
Good! (Happy)

We have:
$$\epsilon = -\d \Phi t = -\d {(B_i A_i + B_o A_o)} t\approx -A_i\d {B_i} t $$
where $A_i$ is the cross sectional area inside the solenoid, $B_i$ is the magnetic field inside the solenoid, $A_o$ is the remaining area outside the solenoid, and $B_o\approx 0$ is the magnetic field outside the solenoid and inside the loop.

This is what we have in a loop outside and around the solenoid, where we neglect the reverse magnetic field, which we can as long as the loop is small enough.

OK, makes sense now. Thanks!
 

Similar threads

Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
1K
Replies
9
Views
6K
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K