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### Question Description

I need help for the solution 7 calculus based EM Problems

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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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