A thin wire of length (2 m) and (1 mm2 ) cross-section area is clamped horizontally between two walls, a weight of (10 kg) is hu
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Answer:
HERE'S MY UNDERSTANDING OF THE DIFFERENCE
Answer:
a) fem = - 0.0103 V, b) the applied field is in a vertical upward direction, the induced current is clockwise.
Explanation:
a) For this exercise we use Faraday's law
fem =
the magnetic flux is
Ф = B. A = B A cos θ
The bold letters indicate vectors, in this case the direction of the magnetic field and the normal to the circumference is parallel therefore the angle is zero and the cos 0 = 1
fem = - B dA / dt
the area of a circle is
A = π r²
l
et's perform the derivative
dA / dt =π 2r
we substitute
fem = - B 2π r
the circumference of a circle is
L = 2π r
we substitute
fem = - B L ( L )
fem =
Let's find the circumference for the 9 s, let's use a direct rule of proportions
If the circumference changes 13cm at t = 1. how much does it change at t=9s
ΔL = 13cm (9s / 1s) = 117cm
the circumference that is
L = Lo - ΔL
L = 167 - 117
L = 50 cm
let's reduce all magnitudes to the SI system
L = 0.50 m
= 0.130 m / s
calculate us
fem = - 1.00 0.50 0.130
fem = - 0.0103 V
b) the electromotive force induced in the opposite direction to the change of the radius and is decreasing with time, the current follows the direction of the decreased voltage therefore the current is induced in the opposite direction to the change of the magnetic flux.
If the applied field is in a vertical upward direction, the induced current is clockwise.