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2 years ago

A wire has an electric field of 6.2 V/m and carries a current density of 2.4 x 108 A/m2. What is its resistivity

Physics

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer:

The resistivity is $\rho = 2.5 *10^{-8} \ \Omega \cdot m$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 6.2 V/m$

The current density is $J = 2.4 *10^{8} \ A/m^2$

Generally the resistivity is mathematically represented as

$\rho = \frac{E}{J}$

substituting values

$\rho = \frac{6.2}{2.4 *10^{8}}$

$\rho = 2.5 *10^{-8} \ \Omega \cdot m$

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A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po

Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, $L_{a} = 1 m$

Frequency, $\vartheta = 1.5 MHz = 1.5\times 10^{6} Hz$

Power transmitted, $P_{t} = 4 W$

Now,

For a monopole antenna:

$\lambda_{a} = \frac{c}{\vartheta}$

where

$\lambda_{a}$ = wavelength transmitted by the antenna

c = speed of light in vacuum

$\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m$

Now,

Since, the value of $\lambda_{a}$ >> $L_{a}$ thus the monopole is a Hertian monopole.

The resistance is calculated as:

$R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}$

$R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega$

$P_{radiated} = P_{t}$

$P_{radiated} = \frac{R}{I^{2}}$

Now, the current I is given by:

$I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A$

5 0

2 years ago

Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.

VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0

2 years ago

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

4 0

2 years ago

13. An aircraft heads North at 320 km/h rel:

AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

$v' = v + v_a$

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

$v_a$ is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

$v_a = -80 km/h$ (since the air is moving from North)

Therefore, we find

$v'=+320 + (-80) = +240 km/h$ (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0

2 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

2 years ago