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3 years ago

13

What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius

0.0200 m?

1 answer:

Nuetrik [128]3 years ago

8 0

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

$R = \frac{mv}{qB}$

here we know that

$q = 1.6 \times 10^{-19}C$

$m = 1.6 \times 10^{-27} kg$

$R = 0.0200 m$

$v = 5 \times 10^5 m/s$

now we have

$B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}$

so we have

$B = 0.25 T$

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3 years ago

Read 2 more answers

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

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3 years ago

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exis [7]

Answer:

A type of telescope that does not require darkness in order to be able to use it is the refracting telescope

Explanation:

A refracting telescope consists of a lens and an eyepiece collects light which is then focused to present a magnified, bright and clear image.

The incident light on a refracting telescope is bent by refraction such that the light is focused to the focal point.

In refracting telescopes, the image is formed by bending light, that is by refraction.

The refracting telescope technology has been applied to binoculars and camera zoom lenses.

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2 years ago

A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358

aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

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3 years ago