A charged particle, with charge +Q, is moving near a current carrying wire with
ID: 2294364 • Letter: A
Question
A charged particle, with charge +Q, is moving near a current carrying wire with the direction of the velocity as shown. What is the direction of the magnetic force on the moving charge due to the wire?
Up and Left
Down and Right
Out of the screen
Into the screen
Left, in the plane of the screen
Down and Left
The force is zero
Up, in the plane of the screen
Right, in the plane of the screen
Down, in the plane of the screen
Up and Right
Two long parallel wires are carrying currents I1 and I2 as shown. I1 lies along the x-axis, at y = 0, and I2 is at y = 23 cm.
If I1 = 4.75 I2, at what point on the y-axis will the total magnetic field be zero?
Give your answer in cm to three digits.
Two long parallel wires are carrying currents I1 = 640 A and I2 = 85 A as shown. I1 lies along the x-axis, at y = 0, and I2 is at y = 3.1 x 10-3 m.
What is the magnitude of the total magnetic field half-way between the two wires?
Give your answer in Tesla to three significant digits.
A current-carrying wire is held stationary near a conducting loop as shown in the picture. The loop and wire lie in the same vertical plane (view the screen or paper with the picture verically) so that if the loop is released if will fall to towards the bottom of the picture.
If the loop is released, what direction will the induced current in the loop flow? Consider a time shortly after the loop is released as it is moving downwards.
Clockwise
Counter Clockwise
There will be no current
This can not be determined
Up and Left
Down and Right
Out of the screen
Into the screen
Left, in the plane of the screen
Down and Left
The force is zero
Up, in the plane of the screen
Right, in the plane of the screen
Down, in the plane of the screen
Up and Right
Explanation / Answer
1) B is to the left
so v into B gives into the page
so Into the screen
2) have to be inbetween the two wires
so B1 = B2 so they cancel
u0 4.75 I2/(2 pi y) = u0 4.75 I2/(2 pi (.23 -y))
4.75/y = 1/(.23 - y)
y=0.19 m
3)
they wills ubtract so
B = B1 - B2 = u0 (I1 - I2)/(2 pi y)
= 4*pi*10^(-7)*(640-85)/(2*pi*3.1E-3/2)
=0.0716 T
4) as it falls it will get further away so there will be less flux
so will put its B into the page
so clockwise current
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