Question

A square coil, enclosing an area with sides 2.0 cm long, is wrapped with 2 500 turns of wire. A uniform magnetic field perpendicular to its plane is turned on and increases to 0.25 T during an interval of 1.0 s. What average voltage is induced in the coil

Answer 1

By using Faraday's law and given values in the question, we determine that the average voltage induced in the square coil with 2,500 turns of wire is -0.25 V.

Step-by-step explanation:

This problem can be solved using the concept of electromagnetic induction and Faraday's law. Given the number of wire turns N = 2500, the area of the coil A = (2.0 cm)^2 = 0.0004 m^2, and the change in magnetic field strength ΔB = 0.25 T occurring in a time period Δt = 1.0 s, we can plug these values into Faraday's law equation:

emf = -N * (ΔΦ/Δt)

Where ΔΦ is the change in magnetic flux. Since the magnetic field is uniform and perpendicular to the plane of the coil, flux Φ = B * A. Therefore, the change in flux ΔΦ = ΔB * A = 0.25 T * 0.0004 m^2 = 0.0001 Wb (Webers).

Substituting the values, we have:

emf = -2500 * (0.0001 Wb / 1.0 s)

Hence, the average voltage (or emf) induced in the coil is -0.25 V. The negative sign indicates that the induced emf opposes the change in magnetic flux, according to Lenz's law.

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Answer 2

The induced voltage in the coil is 0.25 V.

Step-by-step explanation:

It is given that,

Area of a square coil is 2 cm or 0.02 m

Number of turns in the wire is 2500

A uniform magnetic field perpendicular to its plane is turned on and increases to 0.25 T during an interval of 1.0 s.

We need to find the induced voltage in the coil. According to Faraday's law, the induced emf in the coil is given by the rate of change on magnetic flux. So,

`\epsilon=(-d\phi)/(dt)\\\\\epsilon=(-d(NBA))/(dt)\\\\\epsilon=NA(-dB)/(dt)\\\\\epsilon=-2500* (0.02)^2* (0.25)/(1)\\\\\epsilon=-0.25\ V`

So, the induced voltage in the coil is 0.25 V.