Problem Four (Optional) The figure below shows a cross sectional view of four pa

Question

Problem Four (Optional) The figure below shows a cross sectional view of four parallel wires carrying equal currents "i" in the same direction. 2 a. What is the direction of the net magnetic field at the left hand wire, created by the currents in the other three wires? Show your work/reasoning. b. Write an expression for the magnitude of the net magnetic field at the left hand wire, created by the currents in the other three wires? Show your work. c. What is the direction of the magnetic force on the left hand wire caused by the currents in the other three wires? Show your work/reasoning.

Explanation / Answer

Answer for 6:

Principle adopted to solve the above problem is conservation of energy.
The initial energy of the electron is purely kinetic as it is at a great distance from the proton. As it makes entry into the field of proton then the total of potential and kinetic at any distance will be equal to the initial kinetic energy.
Let r be the distance to be found out. Then potential energy (PE) at a point distant r from the proton is - 9x109 x e^2/r . Kinetic energy is four times the initial, as velocity is five twice the initial given one. So the equation becomes
potential energy = kinetic (initial) - 4 kinetic (ini) = - 3 kinetic (ini)
But, kinetic (ini) = 1/2 x m x v2
So 3 Kinetic (ini) = 1.5 x m x v2
Equating PE and - 3 kinetic, we get 9x109 x e2/r = 1.5 x m x v2
Negative sign get cancelled. Now r = 9/1.5m x 10^9 x (e/v)2
Now replacing the values of m (mass of electron) as 9 x 10-31 and e(charge of electron/proton) as 1.6 x 10-19 and v (given) as 3.2 x 105
we can compute the value for the required distance r
as 16.6 x 10-10 m or 16.6 Angstrom.

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