The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.
- The force per unit length between two parallel thin current-carrying
and 
wires at distance ' r ' is given by 
....(1) .
- If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.
A schematic of the information provided in the question can be seen in the image attached below.
From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3
Using equation (1) , we get
I₃ = 2.4 A and the current is pointing in the downward direction
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According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.
Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²
Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
Answer:
0.021 V
Explanation:
The average induced emf (E) can be calculated usgin the Faraday's Law:
<u>Where:</u>
<em>N = is the number of turns = 1 </em>
<em>ΔΦ = ΔB*A </em>
<em>Δt = is the time = 0.3 s </em>
<em>A = is the loop of wire area = πr² = πd²/4 </em>
<em>ΔB: is the magnetic field = (0 - 1.04) T </em>
Hence the average induced emf is:
Therefore, the average induced emf is 0.021 V.
I hope it helps you!
8 miles per hour
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