7 problems for EM

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I need help for the solution 7 calculus based EM Problems

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1. (20 points) A long rectangular resistor of length L is made from a material with conductivity σ. The ends of the resistor are rectangular with width W and height H. The potential difference between the rectangular ends is V. The electric field is constant through the resistor, but its value is not given. a. What is the current that flows from one end to the other? b. What is the resistance? 2. (20 points) Two metal spherical shells are separated by a material with resistivity ρ. The inner metal spherical shell has radius M, and the outer metal spherical shell has radius N. The potential difference between the two shells is V. a. What is the current that flows from the inner shell to the outer shell? b. What is the resistance? 3. (40 points) Test the following hypothetical electric fields to determine if they could be solutions to the wave equation. a. 𝐸 = 𝐴! 𝑒 !"# − 𝐴! 𝑒 !!" b. 𝐸 = 𝐴! cos(𝑘𝑥) + 𝐴! sin(𝜔𝑡) c. 𝐸 = 𝐴! sin(𝑘𝑥 − 𝜔𝑡) − 𝐴! 𝑒 !(!!"!!") d. 𝐸 = 𝐴! 𝑒 !(!"!!") + 𝐴! 𝑒 !!(!"!!") 4. (17.5 points) A steady current  𝐽 = 𝑘𝑠!/! 𝑧 flows down a long cylindrical wire of radius R. (k is a constant.) Find the magnetic field, both inside and outside the wire, using Ampère’s Law. 5. (17.5 points) A slab of current is infinitely long in the y direction. In the z direction, the slab extends from z = -A to z = +A. The current through the slab is 𝐽 = 𝑘𝑧 ! 𝑥 . Find the magnetic field both inside and outside the slab. 6. (17.5 points) A cylindrical, infinitely long region of space of radius R has a magnetic field that can be written as 𝐵(𝑠, 𝜑, 𝑧, 𝑡) = 𝐶𝑠 ! 𝑡 ! 𝑧 . What is the electric field induced inside radius R and outside radius R? 7. (17.5 points) The current to an infinitely long solenoid of radius A is being decreased so that the current can be written as 𝐼 𝑡 = 𝐷𝑒 !! 𝜑. The solenoid has n turns per unit length. What is the electric field induced inside and outside the solenoid? ...
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Tutor Answer

Ace_Tutor
School: New York University

attached is my answer

Problem 1:
a) Since the ends of the resistor are rectangular with width W and height H , we have A  W  H . Let

J and E be the current density and electric field, respectively. The desired flowing current is
determined by

I   J  d A   E  d A
I 

V

V

dA
L
L 
V
V
I
A
 WH
L
L
 HWV
 I
L
 dA 

b) Using Ohm’s law, the resistance is calculated by

V
 HWV
V 
I
L
L
 R V 
 HWV
L
 R
 HW
R

Problem 2:
a) Define J and E be the current density and electric field, respectively. The cross-sectioned area is
calculated by

A  thickness  circumference
 A   M  N   2
 A   M  N
2

2



M  N 
2

Hence, the required current that flows from the inner shell to the outer shell is calculated by

I   J  d A   E  d A
V
V
 dA 
dA
L
L 
V
V
I
A
  N 2  M 2 
L
  2  M  N  / 2
I 

 I

V N  M 



b) Using Ohm’s law, the resistance is calculated by

R

V N  M 
V
V 
I


 R V 
 R



V N  M 


N M

Problem 3:

2 E 1 2 E
for some constant v in order to determine

x 2 v 2 t 2

We are going to check whether E satisfies

whether E is the solution to a wave equation or not.
a) For the given electric field, we have

2 E   

   A1eikx  A2 e t  
2
x
x  x


  ik  A1eixx 
x
  ik  A1eikx  i 2 k 2 A1eikx
2

  1 k 2 A1eikx  k 2 A1eikx
Similarly,

2 E   

   A1eikx  A2 e t  
2
t
t  t


      A2 e t 
t
     A2 e t   2 A2e t
2

Since we cannot have the form

2 E 1 2 E
 2 2 for constant v , it follows E  A1eikx  A2e t cannot be a
2
x
v t

solution to any wave equation.
b) For the given electric field, we have the 2nd derivatives calculated as follow:

2E   

   A1 cos  kx   A2 sin t   
2
x
x  x


   kA1 sin  kx  
x
  k 2 A1 cos  kx 
2E   

   A1 cos  kx   A2 sin t   
2
t
t  t


  A2 cos t  
t
  2 A2 sin t 
We cannot have the form

2 E 1 2 E
for constant v since

x 2 v 2 t 2

2 E 1 2 E

x 2 v 2 t 2
 k 2 A1 cos  kx  
v


k

1
 2 A2 sin t  
2 
v

A2 sin t 
A1 cos  kx 

Hence, E  A1 cos  kx   A2 sin t  cannot be a solution to any wave equation.
c) We have the 2nd derivatives calculated as follow:

2 E    

   A1 sin  kx  t   A2ei  kx t   
2
x
x  x


i  kx t 

  kA1 cos  kx  t    ik  A2e 

x
 k 2 A1 sin  kx  t    ik  A2ei  kx t 
2

 k 2 A1 sin  kx  t   k 2 A2ei  kx t 



 k 2 A1 sin  kx  t   A2e 

i  kx t 



 k 2 E
Similarly,

2 E    

   A1 sin  kx  t   A2ei  kx t   
2
t
t  t


i  kx t 

   A1 co...

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