A 3.00-μC charge is moving with a speed of 4.40 x 104 m/s parallel to a very long, straight wire. The wire is 5.20 cm from the charge and carries a current of 80.0 A. Find the magnitude of the force on the charge.

The magnitude of the force on the charge is 4.0656 x 10⁻⁵ N

Explanation:

Given;

charge on the wire, q = 3.00-μC

speed of the charge, v = 4.40 x 10⁴ m/s

radius from the wire, r = 0.052 m

current in the wire, I = 80.0 A

F = qvB

from biot savart equation, we will determine strength of the magnetic field, B, produced by the current in the wire.

F = qvB = 3 x 10⁻⁶ x 4.4 x 10⁴ x 3.08 x 10⁻⁴ = 4.0656 x 10⁻⁵ N

Therefore, the magnitude of the force on the charge is 4.0656 x 10⁻⁵ N

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can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s