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Physics Electromagnetic Induction

MermaidWonders

Active member
Feb 20, 2018
113
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.
 

MermaidWonders

Active member
Feb 20, 2018
113
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.
 

MermaidWonders

Active member
Feb 20, 2018
113
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. :(
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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?
 

MermaidWonders

Active member
Feb 20, 2018
113
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). :(
 

MermaidWonders

Active member
Feb 20, 2018
113
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.
 

MermaidWonders

Active member
Feb 20, 2018
113
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!