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

In summary, we can use Faraday's law to calculate the induced current in a wire loop outside a solenoid by considering the magnetic field inside the solenoid and neglecting the contribution of the magnetic field outside the solenoid. This is based on the equation $\epsilon = -\d \Phi t = -A_i\d {B_i} t$ where $A_i$ is the cross sectional area inside the solenoid and $B_i$ is the magnetic field inside the solenoid.
  • #1
MermaidWonders
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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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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. :(
 
  • #6
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?
 
  • #7
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). :(
 
  • #8
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?
 
  • #9
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!
 

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

What is electromagnetic induction?

Electromagnetic induction is the process of creating an electric current in a conductor by changing the magnetic field around it. This is done through the movement of the conductor or the changing of the magnetic field.

How does electromagnetic induction work?

Electromagnetic induction works through Faraday's law of induction, which states that a changing magnetic field will induce an electromotive force (EMF) in a conductor. This EMF can then create an electric current in the conductor.

What are the applications of electromagnetic induction?

Electromagnetic induction has many practical applications, such as in generators, transformers, and motors. It is also used in wireless charging, induction stovetops, and metal detectors.

What are the factors that affect electromagnetic induction?

The strength of the magnetic field, the speed at which the magnetic field changes, and the number of turns in the conductor all affect electromagnetic induction. Additionally, the material and properties of the conductor can also impact the induction process.

What are some real-life examples of electromagnetic induction?

Electromagnetic induction is used in everyday devices, such as power plants, electric motors, and generators. It is also utilized in technologies like wireless charging, MRI machines, and induction stovetops. Lightning is also a natural phenomenon that involves electromagnetic induction.

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