Two wires carry currents to the right, where I 1 = 225 A and I 2 = 250 A. A nega
ID: 1462255 • Letter: T
Question
Two wires carry currents to the right, where I1 = 225 A and I2 = 250 A. A negatively charged particle, with charge -55 nC is moving to the right with velocity 4.1x106 m/s.
The wires are a distance d = 5.5x10-3 m apart and the charge is a distance y = 2.5x10-3 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
Here ,
distance of particle from I2 , y2 = 5.5 *10^-3 - 2.5 *10^-3 m
y2 = 3 *10^-3 m
as the magnetic field due to a wire is given as
B = u0*I/(2 *pi *d)
the net magnetic field is
B = B1 - B2
B = 4pi *10^-7/(2*pi) *(225/(2.5 *10^-3) - 250/(3 *10^-3))
solving for B
B = 0.01333 T
Total magnetic force on the particle = q * V X B
Total magnetic force on the particle = -55 *10^-9 * 0.01333 * 4.1 *10^6
Total magnetic force on the particle = - 3.01 *10^-3 N
the force will be downwards
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