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Anarel [89]
3 years ago
5

A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic fi

eld that increases its magnitude at a constant rate from 1.5 T to 5.1 T in 2.0 s. If the coil has a total resistance of 0.20 Ω, what is the magnitude of the induced current?
Physics
1 answer:
zheka24 [161]3 years ago
6 0

Answer:

0.54 A

Explanation:

Parameters given:

Number of turns, N = 15

Area of coil, A = 40 cm² = 0.004 m²

Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T

Time interval, Δt = 2 secs

Resistance of the coil, R = 0.2 ohms

To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:

|V| = |(-N * ΔB * A) /Δt)

|V| = | (-15 * 3.6 * 0.004) / 2 |

|V| = 0.108 V

According to Ohm's law:

|V| = |I| * R

|I| = |V| / R

|I| = 0.108 / 0.2

|I| = 0.54 A

The magnitude of the current in the coil of wire is 0.54 A

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The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

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r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

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