Answer:
=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
Answer:
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.
