I hope this helps. Good luck!

[fredrick08]

Homework Statement

The magnetic field at one place on the earth’s surface is 55 μT in strength and tilted 60°
down from horizontal. A 200-turn coil having a diameter of 4.0 cm and a resistance of 2.0 Ω
is connected to a 1.0 μF capacitor rather than to a current meter. The coil is held in a
horizontal plane and the capacitor is discharged. The coil is then quickly rotated 180° so that
the side that had been facing up is now facing down. Afterward, what is the voltage across
the capacitor? [Hint: use I=dq/dt

Homework Equations

$$\Phi$$=ABcos$$\theta$$
Vc=Q/C
Vl=-L(di/dt)
Vc=-Vl
I=$$\Phi$$/L
$$\Phi$$=N$$\Phi$$per coil

The Attempt at a Solution

ok wat i did was find flux per coil=$$\pi$$r²Bcos$$\theta$$=3.45x10^-8
then flux=3.45x10^-8*200=6.9x10^-6=L(di/dt)=Q/C...
but that can't be right coz i know the answer is 12V... can someone please help me, I've been on this for 4 or hours and its 3am almost... due in tomoz... please someone help...

 

[fredrick08]

please somone help

 

[fredrick08]

no help?

 

[alphysicist]

Hi frederick08,

Homework Statement

The magnetic field at one place on the earth’s surface is 55 μT in strength and tilted 60°
down from horizontal. A 200-turn coil having a diameter of 4.0 cm and a resistance of 2.0 Ω
is connected to a 1.0 μF capacitor rather than to a current meter. The coil is held in a
horizontal plane and the capacitor is discharged. The coil is then quickly rotated 180° so that
the side that had been facing up is now facing down. Afterward, what is the voltage across
the capacitor? [Hint: use I=dq/dt

Homework Equations

$$\Phi$$=ABcos$$\theta$$
Vc=Q/C
Vl=-L(di/dt)
Vc=-Vl
I=$$\Phi$$/L
$$\Phi$$=N$$\Phi$$per coil

The Attempt at a Solution

ok wat i did was find flux per coil=$$\pi$$r²Bcos$$\theta$$=3.45x10^-8

I don't believe this number is right; it looks like you are using the wrong angle. Since the coil is horizontal, you need to find the vertical component of the field to find the flux through the coil.

then flux=3.45x10^-8*200=6.9x10^-6=L(di/dt)=Q/C...

Multiplying by 200 is right; once you find the correct flux per coil this will give you the total flux passing through the coil. (However the result will not equal L(di/dt) or Q/C.)


But you still have more to do. Remember that the magnitude of the induced emf in terms of average values is

$${\cal E}=\frac{\Delta\Phi}{\Delta t}$$

(it's the magnitude because I've dropped the minus sign). So $${\cal E}$$ is the induced emf in volts.

So first you you need the change in the flux $\Delta\Phi$. You have already found the initial flux $\Phi$, so what is $\Delta\Phi$ in this problem (since the coil is flipping over)?


You can use Ohm's law to replace the ${\cal E}$, and also use the hint given in the problem (in terms of averaged values it would be

$$I=\frac{\Delta Q}{\Delta t}$$

You have already shown that you know you are looking for the quantity $Q/C$, so you should be able to get the answer.

but that can't be right coz i know the answer is 12V... can someone please help me, I've been on this for 4 or hours and its 3am almost... due in tomoz... please someone help...