Magnetic Field Problem Urgent HELP PLEASE

[bjm2130]

Magnetic Field Problem! Urgent! HELP! PLEASE!

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harmonphys.info/magnetism%20problems.doc 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.

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Heres what I got so far!

A) I used the Right Hand Rule and I got Into the Page
B) I am not sure but I think you would use B=2E-7(I)/(r) where r the distance from the origin to point P where i get r=1.12m and I=3A and I got B=5.366E-7T but I am wondering if i have to do it again for the 1A current and them add them
C) Positive. right hand rule
D)would you use F=qvB where force equals 2E-7(I)(I)(length of wire)/(r)
E) I got out of the page because the magnetic field is into the page and I'm having a lot of trouble with the reast of part E

 

[DaveC426913]

We do not spoon feed homework here. Show your attempts at a solution and we will try to help.

 

[nlightner]

I understand that you are facing a difficult problem and are in need of urgent help. I will do my best to provide a clear and concise response to your request.

Firstly, it is important to note that the direction of the magnetic field is given by the Right Hand Rule, as you have correctly used. According to the given information, the net magnetic field at point P would be into the page, as the two wires are perpendicular and carry currents in opposite directions.

To determine the magnitude of the net magnetic field, you can use the formula B = 2 X 10^(-7) I/d, where I is the total current and d is the distance from the wire. In this case, the total current would be 4 amperes (3A + 1A), and the distance from the wire carrying 3A would be 0.5m. Plugging in these values, we get B = 8 X 10^(-7) T.

Moving on to part C, the charge on the particle can be determined using the formula F = qvB, where F is the force experienced by the particle, v is its velocity, and B is the magnetic field. Since the force experienced by the particle is given as 10^7 N, and the velocity is 10^6 m/s, we can calculate the charge to be 10^(-1) C.

For part E, we can use the formula F = qE, where E is the electric field. Since the net force on the particle is zero, we can equate the two forces (magnetic and electric) and solve for E. Plugging in the known values, we get E = vB = 10^7 N/C, with a direction out of the page.

In conclusion, the direction of the net magnetic field at point P is into the page, with a magnitude of 8 X 10^(-7) T. The charge on the particle is positive with a magnitude of 10^(-1) C. To make the net force on the particle equal to zero, an electric field of 10^7 N/C, directed out of the page, is required. I hope this helps to clarify the problem and assist you in finding a solution. If you have any further questions or concerns, please do not hesitate to seek additional help.