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

Answer:

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:

$B=\frac{\mu_o I}{2\pi r}$

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

I: current = 850A

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

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

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)