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Citrus2011 [14]
3 years ago
15

A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electrom

agnet, where the magnetic field has a magnitude of B = 0.574 T and is horizontal.
a) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is east?

b) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is south?

c) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is 30.0 degrees south of west?

d) What is the direction of this magnetic force?
Physics
1 answer:
enot [183]3 years ago
8 0

Answer:

a) F = 0.01234N to the south direction

b) F = 0.01234N to the north direction

c) F = 0.01234N

d) North of West

Explanation:

The magnetic force in a magnetic field is given by:

F = BIlsin \theta

Length of the wire, l = 1 cm = 0.01 m

Current flowing through the wire, I = 2.15 A

a) If the magnetic field direction is east

∅ = 90

F = 0.574 * 2.15 * 0.01

F = 0.01234 N to the south direction according to the Fleming's Right Hand Rule

b) If the magnetic field direction is south

The magnitude of the magnetic force remains the same

That is , F = BIL sin 90

F = 0.01234 N to the west

c) If the magnetic field direction is 30 degrees south

The angle between the magnetic field and the length of the wire still remains 90 degrees

Therefore the magnitude of the force still remains 0.01234

F = 0.01234 N

d) the direction of the force is the North of West

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
The energy of a photon of light is __________ proportional to its frequency and __________ proportional to its wavelength.
icang [17]

Answer:

D) directly, inversely

Explanation:

The energy of a photon of light is directly proportional to its frequency and inversely proportional to its wavelength.

Frequency is the number of waves that passes through a point per unit of time.

Wavelength is the is the distance between successive crests or troughs on a wave.

 Mathematically, frequency is related to wavelength and velocity using;

             Energy   =  h x f

                      where h is the Planck's constant

                                  f is the frequency

         

                   Since c  = f ∧

where f is the frequency of the wave

           ∧ is the wavelength of the wave

           c is the speed of light

So;

                 f  = c/∧

  Therefore;

                E  = \frac{h c}{wavelength}

From the equation, we see an inverse relationship between E and wavelength and a direct one with frequency.

           

6 0
3 years ago
Systems distribute heat energy by the circulation of air.
Ulleksa [173]
That seems like a statement more than a question. Where's the question?
6 0
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_____ heat provides energy for thunderstorms and hurricanes. Polar Geothermal Nuclear Latent
NikAS [45]
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4 0
3 years ago
An astronaut is doing repairs on the international Space Station. She throws a hammer into space with a velocity of 3m/s. How mu
Rainbow [258]

Answer:

The force is 15N

Explanation:

The formula is Force= mass × velocity.

From the question mass is 5kg, velocity is 3m/s.

F= 5×3

F= 15Newton.

Therefore the force is 15N.

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