Question

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil oriented for maximum flux. The coil is connected to an uncharged 1.0 μF 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.

Express your answer with the appropriate units.

Answer 1

The voltage across the capacitor is 1.57 V.

Step-by-step explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

`\phi=NBA`

Put the value into the formula

`\phi=10*1.0*10^(-3)*\pi*(0.5*10^(-2))^2`

`\phi=7.85*10^(-7)\ Tm^2`

We need to calculate the induced emf

Using formula of induced emf

`\epsilon=(d\phi)/(dt)`

Put the value into the formula

`\epsilon=(7.85*10^(-7))/(dt)`

Put the value of emf from ohm's law

`\epsilon =IR`

`IR=(7.85*10^(-7))/(dt)`

`Idt=(7.85*10^(-7))/(R)`

`Idt=(7.85*10^(-7))/(0.50)`

`Idt=0.00000157=1.57*10^(-6)\ C`

We know that,

`Idt=dq`

`dq=1.57*10^(-6)\ C`

We need to calculate the voltage across the capacitor

Using formula of charge

`dq=C dV`

`dV=(dq)/(C)`

Put the value into the formula

`dV=(1.57*10^(-6))/(1.0*10^(-6))`

`dV=1.57\ V`

Hence, The voltage across the capacitor is 1.57 V.