Physics, 08.12.2020 16:30 Adithya27

Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 cm. If the current in one wire is 1.80 A and the current in the other wire is 3.45 A running in the opposite direction, determine the magnitude and direction of the force per unit length the wires exert on each other.

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Answer from: mlbaker2381

2.48*10^-5 N/m

Explanation:

The magnitude force per unit length on parallel wires separated by a distance D and currents I1 & I2.

F/l = µ.I1.I2/2πd

Where

µ = 4π *10^-7

I1 = 1.80 A

I2 = 3.45 A

d = 5 cm = 0.05 m

F/l = (4π*10^-7 * 1.8 * 3.45) / (2 * π * 0.05)

F/l = (1.2568*10^-6 * 6.21) / 0.3142

F/I = 7.8*10^-6 / 0.3142

F/l = 2.48*10^-5 N/m

Therefore, the magnitude of the force per unit length is 2.48*10^-5 N/m

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

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Two long current-carrying wires run parallel to each other and are separated by a distance of 5.00 c...

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