A charged particle, with charge +Q, is moving near a current carrying wire with
ID: 2294627 • 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?
The force is zero
Up, in the plane of the screen
Down and Left
Up and Left
Left, in the plane of the screen
Down and Right
Into the screen
Out of the screen
Up and Right
Down, in the plane of the screen
Right, in the plane of the screen
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 = 5 cm.
If I1 = 4.00 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 = 700 A and I2 = 95 A as shown. I1 lies along the x-axis, at y = 0, and I2 is at y = 3.6 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
The force is zero
Up, in the plane of the screen
Down and Left
Up and Left
Left, in the plane of the screen
Down and Right
Into the screen
Out of the screen
Up and Right
Down, in the plane of the screen
Right, in the plane of the screen
Explanation / Answer
1.
Up, in the plane of the screen
2.let this point be y from i1, or it is at (0.05-y) from i2
so for net fiedl to be zero, B1 = B2
u0i1/2pi(y) = u0i2/2pi(0.05-y)
0.05-y/y = 4
4y +y = 0.05
y = 0.05/5 = 1cm from i1 net field is zero
c. Bnet = B1+B2
Bnet = u0/2pid)(i1+i2)
Bnet = (4pi*10^-7/pi*0.0036)( 700+95)
Bnet = 0.08833 T
d. clockwise
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