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AnnZ [28]
3 years ago
8

A single loop of wire with area A carrying a current I has a magnetic moment µ = IA. If you wanted a single loop of wire with an

area equal to the cross-sectional area of the magnet used here to have the same magnetic moment as your permanent magnet, how large a current would you need flowing through the loop?
Physics
1 answer:
viktelen [127]3 years ago
4 0

Answer:

The current would you need flowing through the loop is

I'=\frac{\mu 'I}{\mu }

Explanation:

The magnetic moment is permanent, the area is equal to:

A=\frac{\mu }{I}

The current is:

I'=\frac{\mu '}{A}

Replacing the area value, the current is:

I'=\frac{\mu 'I}{\mu }

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a fast charged particle passes perpendicularly through a thin glass sheet of index of refraction 1.5. The particle emits light i
NeX [460]

The minimum speed of the particle is the Speed of light in glass is c/μ=2×108m/s.

<h3>Why is the refractive index important?</h3>

The higher the refractive index the slower the light travels, which causes a correspondingly increased change in the direction of the light within the material. What this means for lenses is that a higher refractive index material can bend the light more and allow the profile of the lens to be lower.

Refractive index values are usually determined at standard temperature. A higher temperature means the liquid becomes less dense and less viscous, causing light to travel faster in the medium.

To learn more about the refractive index visit the link

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1 year ago
A learner sets up a circuit to determine the conductivity of certain solids. What piece of apparatus is not required for the nex
Lilit [14]
It is D as u dont need a stop watch aft that
4 0
3 years ago
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration
umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

F=qvB     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

F=qvB=ma       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0
3 years ago
Early in the Earth's history, its surface was covered with an ocean of molten rock. Over time, the molten rock cooled to form a
Elanso [62]
It would be the earth crust
3 0
3 years ago
Read 2 more answers
Two long straight parallel lines of charge, #1 and #2, carry positive charge per unit lengths of λ1 and λ2, respectively. λ1 &gt
Stella [2.4K]

Answer:

Somewhere between the two wires, but closer to the wire carrying λ₂

Explanation:

Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².

Electric Fied due to an electric charge is a vector and its direction is  such that  if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)

According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires  are opposite

In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.

As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.

But for points closer to wire with λ₂  ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance  to get equals E and then Ef = 0

3 0
3 years ago
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