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.

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