Two wires carry currents to the right, where I1 = 73.73 A and I2= 110.76 A. A ne

Question

Two wires carry currents to the right, where I1 = 73.73 A and I2= 110.76 A. A negatively charged particle, with charge -49 C is moving to the right with velocity 4.68x106 m/s.

The wires are a distance d = 4.15x10-3 m apart and the charge is a distance y = 0.92x10-3 m from current I1.

What is the total magnetic force on the charge? Forces that are up (towards I2) are positive and forces that are down (towards I1) are negative.

Give your answer in Newtons to at least three significant figures to avoid being counted incorrect due to rounding.

Explanation / Answer

Given I1 = 73.73 A

I2 = 110.76 A

charge q = - 49 C

velocity v = 4.68 * 10^6 m/s

d = 4.15 * 10^-3 m

y = 0.92 * 10^-3 m

The magnetic field due to a current carrying wire is B = muenot * I / (2pi * r)

The magnetic field at the point due to current I1 is

B1 = (muenot * I1) / (2pi * y)

B1 = (4pi * 10^-7 * 73.73) / (2pi * 0.92 * 10^-3)

B1 = 1.6 * 10^-2 T (outward)

The magnetic field at the point due to current I2 is

B2 = (muenot * I2) / (2pi * (d-y))

B2 = (4pi * 10^-7 * 110.76) / (2pi * (4.15-0.92) * 10^-3)

B2 = 0.68 * 10^-2 T (inward)

the net field is B = B1 - B2

B = (1.6 - 0.68) * 10^-2 T

B = 0.92 * 10^-2 T

The magnetic force is

F = q * v * B

F = - 1.6 * 10^-19 * 4.68 * 10^6 * 0.92 * 10^-2

F = 6.889 * 10^-15 N    (upward)-----------------------------------------------Answer

The direction is given by right hand rule

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