Respuesta :
Explanation:
Given that,
Radius of the coil, r = 4.2 cm
Number of turns in the coil, N = 500
The magnetic field as a function of time is given by :
[tex]B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4[/tex]
Resistance of the coil, R = 640 ohms
We need to find the magnitude of induced emf in the coil as a function of time. It is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V[/tex]
Hence, this is the required solution.
Answer:
Explanation:
Radius of the coil, r = 4.2 cm
number of turns, N = 500
resistance in the circuit, R = 640 ohm
The magnetic field is given by
[tex]B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^{4}[/tex]
(a) According to the Faraday's law of electromagnetic induction, the magnitude of induced emf is given by
[tex]e = \frac{d\phi}{dt}[/tex]
magnetic flux, Ф = N x B x A x Cos 0
Ф = N A B
Differentiate both sides
[tex]\frac{d\phi}{dt}=NA\frac{dB}{dt}[/tex]
[tex]\frac{d\phi}{dt}=500\times 3.14 \times 0.042\times 0.042\times \left ( 1.2\times 10^{-2}+4 \times 2.6\times 10^{-5}t^{3}\right )[/tex]
So, the magnitude of induced emf is given by
[tex]e =3.324\times 10^{-2}+28.8 \times 10^{-5}t^{3} V[/tex]
This is the magnitude of induced emf as the function of time.
