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Luda [366]
3 years ago
11

Two points, point A and point B, are situated above a current-carrying wire. Point B is located at a distance R from the wire, w

hich is twice as far from the wire as point A. By what factor is the magnetic field at point A larger or smaller than the magnetic field at point B?
Physics
1 answer:
padilas [110]3 years ago
7 0

Explanation:

It is given that, there are two points situated above a current-carrying wire. Point B is located at a distance R from the wire, which is twice as far from the wire as point A such that,

r_B=R

And r_A=R/2

We know that the magnetic field at a distance R is inversely proportional to the distance from wire as :

B=\dfrac{\mu_o}{2\pi}\dfrac{I}{r}

B_A\propto\dfrac{1}{(R/2)}

B_B\propto\dfrac{1}{(R)}

\dfrac{B_A}{B_B}=\dfrac{1/(R/2)}{1/R}

\dfrac{B_A}{B_B}=\dfrac{2}{1}

B_A=2B_B

So, the magnetic field at point A larger than the magnetic field at point B by a factor of 2. Hence, this is the required solution.

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
Having greater biodiversity in an ecosystem is beneficial to humans because
11Alexandr11 [23.1K]

Answer:

The answer is A

Explanation:

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OD because Boyle’s law specifically states
4 0
3 years ago
A balloon of diameter 20.0 cm is filled with helium gas at 30°C at 1.00 atm. How
bogdanovich [222]

Answer:

N = 3.54 * 10²³ atoms

Explanation:

The formula to apply here is the idea gas law;

PV = nRT  where ;

P= pressure of the gas= 1.013 * 10⁵ Pa

V= volume of the gas = 4/3 * 3.14 *0.15³= 0.01414 m³

n= amount of a substance = ?

R= ideal gas constant= 8.314

T= temperature= 293 K

Applying the values to the formula;

PV = nRT

1.013 * 10⁵ * 0.01414 = n * 8.314*293

n= 1.013 * 10⁵ * 0.01414 / 8.314*293

n= 0.588 moles

1 mole = 6.022 * 10²⁷ atoms/ mole

0.588 moles = 0.588 * 6.022 * 10²⁷

N = 3.54 * 10²³ atoms

8 0
2 years ago
I need help
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Answer:

C

Explanation:

4 0
3 years ago
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