High School

A 121-turn circular coil of radius 2.85 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. Over an interval of 0.179 s, the magnetic field strength increases from 55.1 mT to 97.9 mT.

Find the magnitude of the average emf (\(E_{\text{avg}}\)) induced in the coil during this time interval, in millivolts.

Answer :

Answer:

The magnitude of the average EMF = 73.83 mv

Explanation:

From faradays law of induction, EMF is given as;

EMF = NA(ΔB/Δt)

We are given that;

N = 121 turns

B1 = 55.1 mT

B2 = 97.9 mT

Thus, ΔB = 97.9 mT - 55.1 mT = 42.8 mT

t = 0.179 s

r = 2.85cm = 0.0285 m

Area = πr² = π x (0.0285)² = 0.0025518 m²

Plugging in the relevant values, we can calculate EMF as;

EMF = (121)(0.0025518)(42.8 mT/0.179) = 73.83 mv

Answer:

0.074 V

Explanation:

Parameters given:

Number of turns, N = 121

Radius of coil, r = 2.85 cm = 0.0285 m

Time interval, dt = 0.179 s

Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T

Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T

Change in magnetic field strength,

dB = Bfin - Bin

= 0.0979 - 0.0551

dB = 0.0428 T

The magnitude of the average induced EMF in the coil is given as:

|Eavg| = |-N * A * dB/dt|

Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²

Therefore:

|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|

|Eavg| = |-0.074| V

|Eavg| = 0.074 V

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