Question

4. There is a horizontal wire with conventional current of 850 A moving to the right. (a) Draw how the magnetic field generated by the current would look. (b) Find the strength of the magnetic field created by the wire at point p which is 2.00 cm below the wire. (c) If a proton is moving to the right 2.00 cm below the wire, what are the magnitude and direction of the force felt by the proton?

Answer 1

Step-by-step explanation:

a) Please see the attached image below.

b) The magnitude of the magnetic field generated by the current in a wire is given by:

`B=(\mu_o I)/(2\pi r)`

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 850A

r: distance where B is calculated = 2.00cm = 0.02m

`B=((4\pi*10^(-7)T/A)(250A))/(2\pi(0.02m))=8.5*10^(-3)T=8.5m T`

As you can see in the image, the direction of B points inside the sheet of paper.

c) The magnetic force is given by:

`\vec{F_B}=q\vec{v}\ X\ \vec{B}`

You can assume that the direction of B is in -z direction, that is, -k. Thus, the direction of the proton is in +x direction, or +i. The direction of the magnetic force over the proton is given by the following cross product:

i X (-k) = - (i X k) = -(-j) = j

Then, the direction of FB is in +y direction, that is, upward respect to the wire.

The magnitude of FB will be:

`|\vec{F_B}|=vBsin\theta=(8.5m T)v`

(it is only necessary to replace by the value of v)