460 meters per second, or about 1,000miles per hour.
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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Answer:
(a) 17634.24 Ω
(b) 0.0068 A
Explanation:
(a)
The formula for inductive inductance is given as
X' = 2πFL................... Equation 1
Where X' = inductive reactance, F = frequency, L = inductance
Given: F = 60 Hz, L = 46.8 H, π = 3.14
Substitute into equation 1
X' = 2(3.14)(60)(46.8)
X' = 17634.24 Ω
(b)
From Ohm's law,
Vrms = X'Irms
Where Vrms = Rms Voltage, Irms = rms Current.
make Irms the subject of the equation
Irms = Vrms/X'...................... Equation 2
Given: Vrms = 120 V, X' = 17634.24 Ω
Substitute into equation 2
Irms = 120/17634.24
Irms = 0.0068 A