How Does Electromagnetism Affect a Charged Particle at a Point?

[sherminator]

Homework Statement

The magnitude of the magnetic field in teslas at a distance d from a long straight wire carrying a current I is given by the relation B = 2 X 10 7 I/d. The two long straight wires shown above are perpendicular, insulated from each other, and small enough so that they may be considered to be in the same plane. The wires are not free to move. Point P, in the same plane as the wires, is 0.5 meter from the wire carrying a current of 1 ampere and is 1.0 meter from the wire carrying a current of 3 amperes.
a. What is the direction of the net magnetic field at P due to the currents?
b. Determine the magnitude of the net magnetic field at P due to the currents.

A charged particle at point P that is instantaneously moving with a velocity of 106 meters per second toward the top of the page experiences a force of 10 7 Newtons to the left due to the two currents.
c. State whether the charge on the particle is positive or negative.
d. Determine the magnitude of the charge on the particle.
e. Determine the magnitude and direction of an electric field also at point P that would make the net force on this moving charge equal to zero.

the picture is here
http://www.google.com/search?source...las+at+a+distance+d+from+a+long+straight+wire
first link, first picture in the link.

Homework Equations

B=uoI/2pir
F=ILB
F=qvB
F=qE

The Attempt at a Solution

a) out, because the 3A wire has a stronger force, and I used RHR
b) B=uoI/2pir, subtracted the two B's, and got 2 X 10^-7 T
c) negative b/c of RHR
d) F=qvB=ILB; qv=IL;q=IL/V. not sure what to do from here
e)F=qE=qvB;E=vB=.2N/C

I'm not 100% confident for any of these, so if someone could just confirm/correct these answers, with explanations if wrong, that would be greatly appreciated.

 

[anathema]

Thank you!

a) Your reasoning for the direction of the net magnetic field at P is correct. The magnitude of the magnetic field at P due to the 1A wire is B1 = 2 X 10^-7 T while the magnitude of the magnetic field at P due to the 3A wire is B2 = 6 X 10^-7 T. Using vector addition, the net magnetic field at P is given by B = √(B1^2 + B2^2) = √(4 X 10^-14 + 36 X 10^-14) = 6.082 X 10^-7 T.

b) Your calculation for the magnitude of the net magnetic field at P is correct.

c) Your reasoning for the charge on the particle is correct. Since the force experienced by the particle is to the left, the charge must be negative.

d) To determine the magnitude of the charge on the particle, we can rearrange the equation q=IL/V to get q=mvL/V, where m is the mass of the particle and v is its velocity. We can use the given information to calculate the values of m, L, and V. The mass of the particle can be found using the given velocity and the kinetic energy equation: K = 1/2mv^2. The length of the wire can be calculated using the distance from the wire and the angle between the wire and the particle's velocity. The voltage can be calculated using the given force and the equation F=qE. Once you have all these values, you can plug them into the equation q=mvL/V to find the charge on the particle.

e) The net force on the moving charge is given by F=qvB. Since we want the net force to be zero, we can set this equation equal to zero and solve for the electric field: E = -vB/q. Plugging in the values for v, B, and q that we calculated in part d, we get E = -53.8 N/C, pointing towards the top of the page.

I hope this helps! Let me know if you have any further questions.