Two wires carry currents to the right, where I 1 = 235 A and I 2 = 210 A. A nega
ID: 1462564 • Letter: T
Question
Two wires carry currents to the right, where I1 = 235 A and I2 = 210 A. A negatively charged particle, with charge -55 nC is moving to the right with velocity 4.5x106 m/s.
The wires are a distance d = 4.5x10-3 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
magntic field acting at pint y due to wire carrying current I1 = uoI1/2pi*y
B due to I1 = 4pi*10^-7* 235/ (2pi* 0.5*10^-3)
B due to I1 = 0.094 T (out of the page)
magntic field acting at pint y due to wire carrying current I1 = uoI1/2pi*(d-y)
B due to I2 = 4pi*10^-7* 235/ (2pi* 4*10^-3)
B due to I2 = 0.01175 T (inside the page)
Net magentic field = (0.094-0.01175) T = 0.08225T (out of the page)
magnetic force acting on the charge = q(V * B) = qVB sin a
where a is the angle between Vand B
F = 55*4.5*10^6*0.08225*sin 90
F = 4.59*10^8 N upwards
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