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LUCKY_DIMON [66]
3 years ago
12

In this experiment you will use the Hall Effect to measure the strength of magnetic fields. Which of the following are true stat

ements about the Hall Effect? (Select all that apply.)a.)The value of the voltage across a Hall probe is proportional to the square of the magnetic field in which the probe has been placed.b.)The Hall probe works best when you align the direction of the current flowing in the probe and the direction of the magnetic field to be parallel.c.)The Hall probe that we will use in this lab is made of a semiconductor.d.)The Hall Effect gives rise to a voltage difference across a conductor, perpendicular to the direction of the current flow in the region of magnetic field.
Physics
1 answer:
fgiga [73]3 years ago
8 0

Answer:

c.)The Hall probe that we will use in this lab is made of a semiconductor.

d.)The Hall Effect gives rise to a voltage difference across a conductor, perpendicular to the direction of the current flow in the region of magnetic field.

Explanation:

Hall effect :

 When a magnetic field  applied on the semiconductor in which current is flowing perpendicular to the applied magnetic field then it produce voltage difference perpendicular to the current as well as magnetic field.Magnetic filed and voltage are proportional to each other.By the using of hall effect this is used to measure speed ,position and magnetic field.

So the option (C) and (D) is correct.

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Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

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the relation for the energy and velocity is given by

E = \frac{1}{2}mv^{2}

So, v = \sqrt{\frac{2E}{m}}   .... (1)

Now,

r = \frac{mv}{Bq}

Substitute the value of v from equation (1), we get

r = \frac{\sqrt{2mE}}{Bq}

Let the radius of the alpha particle is r2.

For proton

So, r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}    ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}    ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

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Answer:

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