A coil of wire is connected to an uncharged capacitor in a magnetic field

Click For Summary
A 10-turn coil with a diameter of 1.0 cm and a resistance of 0.20 ohms is connected to an uncharged 1.0 microFarad capacitor and placed in a 1.0 mT magnetic field. When the coil is quickly pulled out of the magnetic field, it induces an electromotive force (emf) due to the change in magnetic flux, calculated to be 7.85 x 10^-7 Wb. The initial voltage across the capacitor is 0 V, as it is uncharged. The discussion highlights the need for a time variable to determine the voltage across the capacitor after the coil is removed from the magnetic field. The lack of a specified time complicates the calculation of the induced voltage.
ANON
Messages
3
Reaction score
0
a coil of wire is connected to an uncharged capacitor in a magnetic field...

Homework Statement


A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.20 ohms is in a 1.0 mT magnetic field
with the coil oriented for maximum flux. The coil is connected to an uncharged 1.0 microFarad capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field. Afterward, what is the voltage across the capacitor?
Hint: Use I=dq/dt to relate the net change of flux to the amount of charge that flows to the capacitor.


Homework Equations


potential difference = -L dI/dt, where L is the inductance
magnetic field, B = Uo NI/l
total magnetic flux = N (magnetic field for each turn) = N (Area)(B) = (Uo N^2 A I) / l
I = V/R
induced emf = |change in magnetic flux/change in time|
V = Q/C

The Attempt at a Solution


Voltage across the capacitor initially is 0 V, since capacitor is uncharged.
I found the total flux to be 7.85 x 10^-7 Wb.
The changing flux induces an emf.

I'm really stuck here. Can someone please help me? Thanks :)
 
Physics news on Phys.org
ANON said:

Homework Statement


A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.20 ohms is in a 1.0 mT magnetic field
with the coil oriented for maximum flux. The coil is connected to an uncharged 1.0 microFarad capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field. Afterward, what is the voltage across the capacitor?
Hint: Use I=dq/dt to relate the net change of flux to the amount of charge that flows to the capacitor.

Homework Equations


potential difference = -L dI/dt, where L is the inductance
magnetic field, B = Uo NI/l
total magnetic flux = N (magnetic field for each turn) = N (Area)(B) = (Uo N^2 A I) / l
I = V/R
induced emf = |change in magnetic flux/change in time|
V = Q/C

The Attempt at a Solution


Voltage across the capacitor initially is 0 V, since capacitor is uncharged.
I found the total flux to be 7.85 x 10^-7 Wb.
The changing flux induces an emf.

I'm really stuck here. Can someone please help me? Thanks :)
no time is given? if you're given a time then you have \frac{d\phi _B}{dt} and a voltage
 
No time is given.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
152
Views
7K
  • · Replies 4 ·
Replies
4
Views
3K