Complete Question
A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
The value is
Explanation:
From the question we are told that
The cross-sectional area is
The number of turns is
The initial time is t = 0s
The current on the solenoid is
The number of turns of the secondary winding is
Generally At I = 3.2 A
=>
=>
Generally induced emf is mathematically represented as
The electric force between two charges is:
F = (9 x 10⁹) Q₁ Q₂ / D²
F is the force, in Newtons
Q₁ and Q₂ are the two charges, in Coulombs
D is the distance between them, in meters
For these two particles:
F = (9 x 10⁹) (0.35) (0.35) / (1)²
F = (9 x 0.35 x 0.35 x 10⁹) / (1)
<em>F = 1.10 x 10⁹ Newtons</em>
Thatsa lotta force . . . like <em>124 thousand tons</em> !
The reason it's so big is because the charges in this question are so big ... 0.35 Coulombs each. 1 Coulomb is a huge amount of charge.
Each of the particles feels the same force, pushing it away from the other particle. (The electric force between two charges is always the same in both directions, just like the gravitational force between two masses.)
The answer is C.
they run east to west which makes them parallel to the equator and they measure distances north and south.
I hope this helps.