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Lana71 [14]
3 years ago
5

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

lue of the magnetic field in the wave
Physics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

7.78x10^-8T

Explanation:

The Pointing Vector S is

S = (1/μ0) E × B

at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,

S = (1/μ0) E B

where S, E and B are magnitudes. The average value of the Pointing Vector is

<S> = [1/(2 μ0)] E0 B0

where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)

Also at any instant,

E = c B

where E and B are magnitudes, so it must also be true at the instant of peak values

E0 = c B0

Substituting for E0,

<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²

Solve for B0.

Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)

= 7.79 x10 ^-8 T

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The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of
PIT_PIT [208]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

\Phi_B=S\cdot B=SBcos\alpha        (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT

The strength of the magntetic field is 4.1mT

4 0
3 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

F_c=F_e=ma_c      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

The radius of the circular trajectory is 0.22m

5 0
3 years ago
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