A 8.0 cm diameter wire coil is initially oriented so that its plane....?

Question

is perpendicular to a magnetic field of 0.63 T pointing up. During the course of 0.25 s, the field is changed to one of 0.25 T pointing down. What is the average induced emf in the coil? mV


Can anyone offer any help for this one??

Answer 1

From maxwell's equations in integral form

∫E*ds = ∫ ∂B/∂t * dA

The line integral of electric field around an area containing a time-varying magnetic field is equal to the rate of change of the magnetic field integrated over the area.

You have a simple geometric arrangement: the field is uniform inside the coil, so the area integral is simply B*A = B*π*r^2. You can figure the time rate of change of B since the field goes from 0.63T to - 0.25T, or 0.88T change in 0.25 sec. Therefore the right-hand side integral is

(0.88T/0.25sec)*π*(4.0cm)^2

....................b
The integral ∫E*ds is the potential between a and b from field E
...................a

Then V = (0.88T/0.25sec)*π*(4.0cm)^2

You can compute this, but watch the units (you have sec, teslas and cm--convert all to consistent units).

Answer 2

Pi is the ratio of the circumference to the diameter. C= pi * d C = pi * 22.32 discover the circumference. Circumference = 26 * pi After that multiply the fee in line with metre by the circumference. fee = 26.fifty six * circumference