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Kobotan [32]
3 years ago
11

A cylindrical conductor with a circular cross section has a radius a and a resistivity p and carries a constant current I. (Take

the current to be coming out of the page when the cross-sectional view of the conductor is in the plane of the page.) For the following questions, express your answer in terms of the variables I,a,p, and appropriate constants (u0,e0, and pi).a) what is the magnitude of the electric field vector E at a point just inside the wire at a distance a from the axis?b) What is the magnitude of the magnetic field vector B at the same point?c) What is the magnitude of the Poynting vector S at the same point?d) What are the directions of these three vectors?
Physics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

Explanation:

a)

Using Ohms Law

R= \rho \frac{l}{\pi a^2 }
\\V = IR = E l = I \rho \frac{l}{\pi a^2 }
\\E = \frac{I \rho}{\pi a^2 }
\\E = J \rho


Where J is the current density J = \frac{I}{\pi a^2 }


and the direction of E is the same as the direction of the current. Since J is uniform throughout the conductor E = \rhoJ just inside at a radius a (and anywhere else).

b)

Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without displacement current

\oint B .dl = B 2 \pi a = \mu_{o} I


such that

B = \frac{\mu_{o} I}{2 \pi a }


and by the right hand rule, since the current is going to the right, the magnetic field is circling around the conductor such that it’s pointing out of the page at the top and into the page at the bottom.

c)

The Poynting vector is given by

S = \frac{1}{ \mu_o} |E \textrm{x}B| = \frac{\rho I^2}{2 \pi^2 a^3}


and by the right hand rule it’s always pointing in towards the center of the conductor.

d)

Note: directions of these three vectors are mentioned along with their magnitudes in above 3 parts a , b and c

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Answer:a

Explanation:

8 0
3 years ago
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.4 cmcm and a current of 12 AA. The bigger
Free_Kalibri [48]

Answer:

Explanation:

Given that,

Current in loops are

i1 = 12A

i2 = 20A

The loops are 3.4cm apart

The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop

Magnetic Field is given as

B= μoi/2πr

Where,

μo is a constant = 4π×10^-7 Tm/A

r is the distance between the two wires

i is the current in the wires

B is the magnetic field

NOTE

Field due to large loop should be equal to the smaller loop.

B1 = B2

μo•i1 / 2π•r1 = μo•i2 / 2π•r2

Then, μo, 2π cancels out, so we have

i1 / r1 = i2 / r2

Make r2 subject of formula

i1•r2 = i2•r1

r2 = i2•r1 / i2

r2 = 20×3.4/12

r2 = 5.67cm

The radius of the bigger loop is 5.67cm.

4 0
3 years ago
The four fundamental types of organic molecules are nucleic acids, proteins, lipids, and ________.
notka56 [123]
The other of the four organic molecules is carbohydrates. 
All four of them are:
Nucleic Acids (DNA and RNA stuff)
Proteins
Lipids
& Carbohydrates
4 0
3 years ago
Read 2 more answers
The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the
krek1111 [17]

Answer:

Explanation:

The force exerted in a magnetic field is given as

F = q (v × B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = -1.602 × 10^-19 C.

Then note that,

V×B is the cross product of the speed and the magnetic field

Then,

F = q (V×B)

F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)

Note

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]

F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]

F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = -1.602 × 10^-19(16•j - 32•i)

F = -1.602 × 10^-19 × ( -32•i + 16•j)

F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j

Then, the x component of the force is

Fx = 5.126 × 10^-18 N

Also, the y component of the force is

Fy = -2.563 × 10^-18 N

8 0
3 years ago
Helpp pls
Kipish [7]

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

5 0
3 years ago
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