Two wires carry currents to the right, where I 1 = 255 A and I 2 = 240 A. A nega
ID: 1461029 • Letter: T
Question
Two wires carry currents to the right, where I1 = 255 A and I2 = 240 A. A negatively charged particle, with charge -40 nC is moving to the right with velocity 6.9x106 m/s.
The wires are a distance d = 4.5x10-3 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 from d1 , y = 2.5 *10^-3 m
distance from d2 , y2 = 4.5 *10^-3 - 2.5 *10^-3
y2 = 2 *10^-3 m
magnetic field at the point P
BP = u0*I1/(2*pi*y) - u0*I2/(2*pi*y2)
BP = 4pi *10^-7/2pi * (255/(2.5 *10^-3 ) - 240/(2 *10^-3 ))
BP = -0.22 T
magnetic force = BP * v * q
magnetic force = 0.22 * 6.9*10^6 * 40 *10^-9
magnetic force = 6.06 *10^-3 N
the magnitude of magnetic force is 6.06 *10^-3 N
as magnetic force = q * V X B
using right hand rule ,
the direction of magnetic force is upwards
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