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poizon [28]
3 years ago
14

A thin slab of Germanium is used as a Hall Effect probe. How would you orient a magnetic field to make the side facing out of th

e page be at a positive voltage with respect to the opposite side facing into the page
Physics
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

the magnetic field must go in a direction parallel to the page perpendicular to the current.

Explanation:

The Hall effect is the voltage produced by the movement of electrons due to the effect of electric and magnetic fields in a material

            F = eE + v x B

The electric field goes in the direction of the current that is opposite to the direction of the electrons, therefore the magnetic force must be perpendicular to it.

Therefore, if the current goes in a direction parallel to the page, in the x direction, the magnetic field must be perpendicular to it if we use the rule of the right wizard,

thumb points in the direction of E, x axis parallel page

The fingers extended should go parallel to the page in the direction and up

The palm is the direction of the Force, where the voltage will be produced points out the page, this is for positive charges, as in germanium the charges are negative, the real force goes into the page.

Therefore the electrons accumulate on the inside of the page and the voltage is negative in this part.

Therefore the voltage is positive on the outside of the sheet. In conclusion the magnetic field must go in a direction parallel to the page perpendicular to the current.

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A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
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Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

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observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

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now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

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total distance

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3 years ago
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Flura [38]

Answer:

about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

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