Suppose that a 0.375 m radius, 500 turn coil produces an average emf of 11950 V when rotated one-fourth of a revolution in 4.12 ms, starting from its plane being perpendicular to the magnetic field. show answer No Attempt What is the peak emf generated by this coil, in volts

B = 0.223T

Explanation:

step 1:

The equation for induced emf is:

emf = N ΔФ/Δt ..............equ(1)

where N = number of turns of the coil

ΔФ = change in magnetic flux of the  coil

Δt = time interval

Magnetic flux is given by

Ф = BA cosθ

where B = magnetic field

A = area

θ = Angle between the normal to the plane of the coil and the magnetic field

Area of teh circular coil is constant and also it is given that the magnetic field is uniform. which implies B and A  are constant so that the change in magnetic flux is due to the change in θ only

The equation for ΔΦ = BA(Δcosθ).............equ(2)

since the coil rotates through one fourth revolution the value of θ changes from 0° to 90°

The value of (Δcosθ) is found as

Δcosθ = cos90° - cos0°

=0-1

which is -1

put the above value in equation (2)

ΔΦ = -BA

step 2:

put the above equation in equation (1)

emf = -N (-BA)/Δt

=NBA/Δt

B = (emf)Δt/NA

The area of the circular coil is given by

A = πr²

where r is the radius of the coil

A = π(0.375m)²

A = 0.442m²

substitute 11950v for emf, 4.12 ms for Δt, 500 for N and 0.442m² for A in equation (3) to find the magnetic field B

B = = 0.223 T

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