Answer:
(a) 0.204 Weber
(b) 0.22 Volt
Explanation:
N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.
(a) Magnetic flux, Ф = N x B x A
Ф = 100 x 0.0650 x 3.14 x 0.1 0.1
Ф = 0.204 Weber
(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s
dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s
induced emf, e = N dФ/dt
e = N x A x dB/dt
e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
It is given that
Number of the turns N = 100,
Radius of the coil r = 10 cm = 0.1 m,
The magnetic field B = 0.0650 T
Since the angle is 90 degrees with the plane of the coil, so [tex]\Theta[/tex]= 0 degrees with the normal coil.
(a) The Magnetic flux, will be given as
[tex]\rm\phi =N\times B\times A[/tex]
By putting the values
[tex]\phi =100\times0.0650\times3.14\times 0.1\times0.1[/tex]
[tex]\phi=0.204 \rm \ weber[/tex]
(b) The emf induced will to be given as
[tex]B_1=0.0650T\ \ B_2=0.1T\ \ dt=0.5s[/tex]
[tex]\dfrac{dB}{dt} =\dfrac{B_2-B_1}{dt} =\dfrac{0.1-0.650}{0.5} =0.07\dfrac{T}{s}[/tex]
induced emf,
[tex]e=N\dfrac{d\phi}{dt}[/tex]
[tex]e=N\times A \times \dfrac{dB}{dt}[/tex]
[tex]e=100\times 3.14\times0.1\times0.1\times0.07=0.22V[/tex]
Thus
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
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