A10-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.

The voltage across the capacitor is 1.57 V.

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

Put the value into the formula

We need to calculate the induced emf

Using formula of induced emf

Put the value into the formula

Put the value of emf from ohm's law

We know that,

We need to calculate the voltage across the capacitor

Using formula of charge

Put the value into the formula

Hence, The voltage across the capacitor is 1.57 V.

When heat rises, the liquid expands from a bowl or bulb into the empty area, climbing up the tube. When the temperature falls, the liquid contracts and goes back down. Liquid thermometers often include both Celsius and Fahrenheit temperature scales, which are displayed on either side of the tube.

Explanation:

Brainliest please