A 34-turn circular coil of radius 5.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 0t + 0.040 0t2, where B is in teslas and t is in seconds. Calculate the induced emf in the coil at t = 6.00 s.

E = 83.7 mV

Explanation:

Let's use the Faraday's law and find the induced emf in the coil:

E = - (dΦ_B) / dt,

Where;

E is the emf generated between the ends of the coil,

The magnetic flux through the coil is Φ_B=NBAcosθ N is the number of turns of the coil, B is the magnetic field,

A=πr² is the cross-sectional area of the coil

r is the radius of the circular coil,

θ is the angle between the magnetic field and the normal to the plane of the coil (since the magnetic field directed perpendicularly to the plane of the coil, θ=0^°).

Thus, E = = NBAcos0°=NBA

Then, we get:

E = (dΦ_B) / dt = d (NBA) / dt = Nπr²d/dt (0.01t+0.04t²)

This gives; E = Nπr² (0.01+0.08t),

At (t = 6 s);

E = 34π (0.05²) (0.01 + (0.08x6) = 0.13085V = 130.85 mV.

=0.1308V

=130.8mV

Explanation:

Given that,

34 - turn coil

radius = 5.00 cm

resistance = 1.00 Ω

B = 0.0100t + 0.0400t²

T = 6.00s

Calculate the induced emf in the coil

lƩl = ΔφB/Δt = N (dB/dt) A

differentiate B = 0.0100t + 0.0400t² in respect to time

=N[d/dt (0.0100t + 0.0400 t^2) A

= 0.0100 + 0.0400 (2) t

= 0.0100 + 0.0800t

=34 (0.0100 + 0.0800 (6.00)) [π (0.05) ^2]

= 34 (0.0100 + 0.48) [π (0.0025) ]

=34 (0.49) [π (0.0025) ]

=0.1308V

=130.8mV