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3 years ago

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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?

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

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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.

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Answer:

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Explanation:

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Answer:

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Explanation:

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