A rectangular coil of 100 turns with dimensions 0.10\text{ m} by 0.20\text{ m} is position… — JAMB UTME

A rectangular coil of $100$ turns with dimensions $0.10\text{ m}$ by $0.20\text{ m}$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . The coil is rotated through $90^\circ$ about an axis perpendicular to the field lines in a time interval of $0.040\text{ s}$ , bringing its plane parallel to the magnetic field. What is the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts?

Answer: 25 V

Answer

The magnitude of the average induced electromotive force in the coil is

25\text{ V}

According to Faraday's law, the induced electromotive force magnitude is

E = N \frac{\Delta \Phi}{\Delta t}

. The initial flux through each turn is

\Phi_1 = B A = 0.50\text{ T} \times 0.020\text{ m}^2 = 0.010\text{ Wb}

. When rotated parallel to the field, the final flux

\Phi_2

0\text{ Wb}

, so

\Delta \Phi = 0.010\text{ Wb}

. Substituting

N = 100

and

\Delta t = 0.040\text{ s}

yields an induced e.m.f. of

100 \times \frac{0.010}{0.040} = 25\text{ V}

Step-by-Step Solution

Calculate the cross-sectional area of the rectangular coil.

A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2

The surface area is required to find the magnetic flux passing through each turn.

Determine the change in magnetic flux through one turn of the coil.

Initial flux

\Phi_1 = B A = 0.50 \times 0.020 = 0.010\text{ Wb}

; Final flux

\Phi_2 = 0\text{ Wb}

; Change

\Delta \Phi = 0.010\text{ Wb}

When perpendicular to the magnetic field, maximum flux links the coil. Rotating it parallel reduces the flux linking the coil to zero.

Apply Faraday's law of electromagnetic induction to calculate induced e.m.f.

E = N \frac{\Delta \Phi}{\Delta t} = 100 \times \frac{0.010\text{ Wb}}{0.040\text{ s}} = 25\text{ V}

Faraday's law states that the induced e.m.f. magnitude equals the rate of change of total magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic