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Neporo4naja [7]
3 years ago
15

What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the lef

t and the bottom wire has a current of 2 A to the right?
A) Out of the page
B) into the page
C) Left
D) The magnetic fields cancel out at that position
Physics
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

Use the B-field formula:

d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \^r}{r^2}

The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

(-\^x) \times (-\^y) = +\^z for the top wire.

(+\^x) \times (+\^y) = +\^z for the bottom wire.

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Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo
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Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy (E_{k}) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        E_{k} = h(f − f₀)

                                        E_{k} = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        E_{k} = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy E_{k} = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from E_{k} = E - ϕ ;

                                        ϕ = E - E_{k}

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

4 0
3 years ago
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