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3 years ago

6

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

is region, it will be..??
deflected toward bottom of the screen
deflected out of the plane of screen
deflected toward top of the screen
deflected into plane of screen
unaffected in its direction of motion

1 answer:

Natali5045456 [20]3 years ago

7 0

Answer:

deflected toward top of the screen

Explanation:

We can find the direction of the force acting on the proton by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: it gives the direction of the force --> therefore, it will be upward

And since the charge is positive (it is a proton), we don't have to revers the direction: so, the force on the proton points upward.

Therefore, the correct answer will be

deflected toward top of the screen

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Answer:

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R is proportional to the length of the wire:

R ∝ length

R is also proportional to the inverse square of the diameter:

R ∝ 1/diameter²

The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.

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Scale factor due to changed diameter:

k₂ = 1/(0.1/0.26)² = 6.76

Multiply the original resistance by these factors to get the new resistance:

R = R₀k₁k₂

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$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

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$\theta)$ the angle between incident phhoton and the scatered photon.

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Now, let's find the angle that will produce a fourth of this maximum value found in (3):

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$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

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Finding $\theta$ :

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$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

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