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bagirrra123 [75]
4 years ago
15

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?

Physics
1 answer:
nata0808 [166]4 years ago
6 0

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)

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-54.12 V

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The change in electrostatic potential energy \Delta U, of one point charge q is defined as the product of the charge and the potential difference.

\Delta U=qV\\V=\frac{\Delta U}{q}\\V=\frac{4.60*10^{-4}J}{-8.50*10^{-6}C}\\V=-54.12 V

5 0
4 years ago
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Answer:

W = 1.06 MJ

Explanation:

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- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

                                             F = p*V*g

                                             F = 1000*(2*0.5*x*8*dx)*g

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- Where, s is the distance from top of hose to the differential volume:

                                             s = (5 - x)

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                                            dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

                                           W = 78400*(2.5*x^2 - (1/3)*x^3)

                                           W = 78400*(2.5*3^2 - (1/3)*3^3)

                                           W = 78400*13.5 = 1058400 J

 

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Answer:

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3 0
3 years ago
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