This is the last question for a final exam preparation sheet and I can't figure out the answer. Any help would be greatly appreciated. Thank you.
- One very long wire carries current 10.0 A to the left along the x axis. A second very long wire carries current 75.0 A to the right along the line
(y = 0.280 m, z = 0).
(a) Where in the plane of the two wires is the total magnetic field equal to zero? (along the y axis)
(b) A particle with a charge of -2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.
(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.
(a)
Magnetic field of the long current carrying wire according to Biot-Savert Law is
B=μₒI/2•π•a
The desired point is separated by distance “x” from the 1st current and by distance (d+x) from the 2nd current (along y-axis).
μₒI1/2•π•x=μₒI2/2•π•(d+x),
10•(0.28+x)=75•x
2.8+10x=75 x.
65x=2.8
x=0.431 m.
(b) Magnetic force (Lorentz force) is
F=qvB sinα.
Since the motion of the particle is along the straight line, then F=0
(c) Electric field has to be applied along the direction of the particle motion (E↑↑ῡ, E↑↓ῡ)
thank you Elena!
For people of the future, Elena's answer for part B is wrong. It's better to use the equation
F = q*v(cross)B
where B is the sum of the B field created by the 50 A wire and the 30 A wire.
------------------------> 50A
|
| .18m
|
------> v = 150E6 m/s
|
| .1m
<-------------------- 30A
B_tot = mu/2pi * (50/.18 (-k) + 30/.1 (-k))
= 1.2E-4 (-k) Teslas
Then you take the cross product and eventually get F = -.035j N
To find the answers to these questions, we can use the principle of superposition and apply the right-hand rule for calculating the magnetic field and force. Let's break down each question.
(a) Where in the plane of the two wires is the total magnetic field equal to zero?
To determine where the total magnetic field is zero, we need to consider the magnetic fields created by each wire separately. The magnetic field created by a long wire can be calculated using the formula:
B = (μ0 * I) / (2πr)
Where B is the magnetic field, μ0 is the permeability of free space (4π x 10^-7 T∙m/A), I is the current, and r is the distance from the wire.
For the wire carrying a current of 10.0 A to the left along the x-axis, the magnetic field will point in the positive y-direction. By applying the right-hand rule (wrap your fingers around the wire with your thumb pointing in the direction of the current), we can determine that the magnetic field due to this wire will be in the positive y-direction.
For the wire carrying a current of 75.0 A to the right along the line (y = 0.280 m, z = 0), the magnetic field will point in the negative y-direction. Similarly, applying the right-hand rule in this case, we find that the magnetic field due to this wire will be in the negative y-direction.
Since the magnetic fields due to the two wires are in opposite directions, there must be a point in between the wires where they cancel each other out, resulting in a total magnetic field equal to zero along the y-axis.
To determine the exact location, we can set up an equation using the magnetic field formula and solve for the position (x-coordinate) at which the magnetic fields add up to zero.
(b) A particle with a charge of -2.00 µC is moving with a velocity of 150 Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle.
To calculate the magnetic force acting on a charged particle moving in a magnetic field, we can use the formula:
F = q * (v x B)
Where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field.
In this case, since the charge of the particle is negative, the force will be opposite in direction to the cross product of velocity and magnetic field.
Given the charge, velocity, and the magnetic field (which we can calculate using the magnetic field formula), we can substitute these values into the formula to find the magnitude and direction of the magnetic force.
(c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.
If a uniform electric field is applied to allow the particle to pass through the region undeflected, the electric force acting on the particle should balance the magnetic force. Since the electric force can be calculated using the formula:
F = q * E
Where F is the electric force, q is the charge of the particle, and E is the electric field.
We can rearrange the formula to solve for the electric field:
E = F / q
By substituting the value of the magnetic force acting on the particle (calculated in part (b)) and the charge of the particle, we can find the required vector electric field that would balance the magnetic force and keep the particle undeflected.