A 0.130 m radius, 485-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.

B = 1.619 T

Explanation:

We are given;

Radius of coil, r = 0.13 m

Number of turns; N = 485-turn

Time of revolution, t = 4.17 ms = 0.00417 s

Average induced emf = 10,000

The equation for the for the induced emf is given by;

EMF = -N(ΔФ/Δt)

where;

N is number of turns of coil

ΔФ is change in magnetic flux

Δt is change in time interval

The change in magnetic flux is given by the formula ;

ΔФ = BAcosθ

where;

B is the magnetic field strength

A is the area of the coil = πr²

θ is the angle between the normal to the plane and the magnetic field,

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

The value of Δcosθ is found as;

Δcosθ = cos90 - cos0 = 0 - 1 = - 1

Thus,

ΔФ = BAcosθ = -BA

And so, EMF = -N(-BA/Δt) = NBA/Δt

Making B the subject, we have;

B = ((EMF)Δt)/NA

A = πr² = π x 0.13² = 0.0531 m²

Thus, B = ((10000•0.00417)/(485•0.0531) = 1.619 T

V = NBA/t

B = (Vt)/NA

B = (10,000 x 0.00417) / (485 x π x 0.13²)

B =1.619 T

Thus, the magnetic field strength needed is 1.619 T

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s