Two wires carry currents to the right, where I 1 = 255 A and I 2 = 240 A. A nega

Question

Two wires carry currents to the right, where I1 = 255 A and I2 = 240 A. A negatively charged particle, with charge -30 nC is moving to the right with velocity 4.1x106 m/s.

The wires are a distance d = 4.0x10-3 m apart and the charge is a distance y = 0.5x10-3 m from current I1.

What is the total magnetic force on the charge? Give your answer in Newtons to three significant figures (NOT three decimal places, but digits). Forces that are up (towards I2) are positive and forces that are down (towards I1) are negative.

Explanation / Answer

Total Magnetic field acting on charge is Bnet = B1 due to wire1 - B2 due to Wiire2

magnetic field B1 = uo I1/(2 pi d1)

where uo is constant = 4pi e -7

I 1 is current

d is distance from Wire

so

B1 = (4*3.14e -7 * 255)/(2*3.14 * 4 e -3) = 0.01275 Tesla

B2 = uo I2/(2pi (d-y) = 4*3.14 e -7* 240/(2*3.14* (4-0.5) *1e -3)

B2 = 0.0137 T

so Bnet = 0.01275 - 0.0137 = 0.00095 T

so Now Magnetic force F = qVB

where V is speed , q is charge and B is Net magnetic field

so

F = 30 e -9 * 4.1 e 6 * 0.00095

F = 1.16 *10^-4 N

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