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kotykmax [81]
3 years ago
14

An experimentalist fires a beam of electrons, creating a visible path in the air that can be measured. The beam is fired along a

direction parallel to a current-carrying wire, and the electrons travel in a circular path in response to the wire's magnetic field. Assuming the mass and charge of the electrons is known, what quantities would you need to measure in order to deduce the current in the wire and the magnetic field due to that current?
Physics
1 answer:
gtnhenbr [62]3 years ago
4 0

Answer:  

Velocity of the electron in the beam.

Radius of the circulating electrons due to the magnetic field.

Explanation:

We have a Mathematical expression for the force on a moving charge in a magnetic field as:

F=q.v.B.sin \theta ...........................(1)

where:

q= charge on the particle in coulomb

v= velocity of the charge projected into the magnetic field

B= intensity of the magnetic field in tesla

\theta= angle between the velocity and direction of magnetic field

For the forces on rotating mass we have the formula:

F=m.\frac{v^2}{r}..........................................(2)

where:

m= mass of the charged particle

v= velocity of projection of charge into the magnetic field

r= radius of the path traced  by the charge in the magnetic field

From eq. (1) and (2) we can calculate the magnetic field .

Now,

Using Ampere's Law we have:

B = \frac{\mu_0 .I}{2 \pi r}

where:

I= current in the wire

\mu_0= The permeability of free space.

r= radial distance from the current carrying wire( in this case it is same as the radius of the circular path)

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A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete
NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}

From Rayleigh criteria

sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm

The focal length of the camera's lens is 300cm

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3 years ago
Is condensation adding thermal energy
Anna35 [415]
I believe it is, since the heat causes the water to evaporate and cause condensation,
3 0
3 years ago
A person's speed around the Earth is faster at the poles than it is at the equator.
VashaNatasha [74]

Answer:

no

Explanation:

it is faster at the equator

6 0
2 years ago
How long must a simple pendulum be if it is to make exactly one swing per four seconds? (That is, one complete oscillation takes
cricket20 [7]

Answer:

Length of the pendulum will be 3.987 m

Explanation:

We have given time period of the pendulum T = 8 sec

Acceleration due to gravity g=9.81m/sec^2

We have to find the length of the simple pendulum

We know that time period of the simple pendulum is given by

T=2\pi \sqrt{\frac{l}{g}}

8=2\times 3.14 \sqrt{\frac{l}{9.81}}

l=3.987m

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3 0
3 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
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