Question

A 121 turn 121 turn circular coil of radius 2.85 cm 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 0.179 as, the magnetic field strength increases from 55.1 mT 55.1 mT to 97.9 mT 97.9 mT. Find the magnitude of the average emf E avg Avg induced in the coil during this time interval, in millivolts.

Answer 1

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 2

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