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3 years ago

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A proton having an initial velocity of 18.1 Mm/s enters a uniform magnetic field of magnitude 0.330 T with a direction perpendic

ular to the proton's velocity. It leaves the field-filled region with velocity -18.1 Mm/s.
(a) Determine the direction of the magnetic field.

(b) Determine the radius of curvature of the proton's path while in the field.
m

(c) Determine the distance the proton traveled in the field.
m

(d) Determine the time interval for which the proton is in the field.
ns

1 answer:

horrorfan [7]3 years ago

5 0

Explanation:

Below is an attachment containing the solution.

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2 years ago

The force vector A has components Ax = 5 N, Ay = -3 N. The force vector B has

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2 years ago

A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field

stepladder [879]

Answer:

$\vec{E} = 1.2\times 10^5(-i)$

Explanation:

Given that

Speed ,v= 2 x 10⁵ m/s ( - y direction)

B= 0.6 T (- z direction)

The resultant force on the proton given as

$\vec{F}=q.\vec{E}+ q.(\vec{v}\times \vec{B})$

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F = 0

$0=q.\vec{E}+ q.(\vec{v}\times \vec{B})$

$0=\vec{E}+ (\vec{v}\times \vec{B})$

$0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)$

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2 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

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$\rho=\frac{m}{V}$

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

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3 years ago