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Natalija [7]
3 years ago
9

If the velocity of a proton is straight up (thumb pointing up) then RHR2 shows that the force points to the left. What would the

direction of the force be if the velocity were
Physics
1 answer:
diamong [38]3 years ago
6 0

Answer:

a) to the right

b) up

c) down

d) there will be no force on the proton

e) there will be no force on the proton

Explanation:

The complete question is

If the velocity of a proton is straight up(thumb pointing up) then RHR2 (right hand rule) shows that the force points to the left. What would the direction of the force be if the velocity were a)down, b)to the right, c)to the left, d)into the page, and e)out of the page?

The right hand rule for a positive charge states that...

Hold the thumb of the right hand at right angle to the rest of the fingers, and the rest of the fingers parallel to one another, and pointing away from the body. If the thumb shows the velocity of a positive charge in a magnetic field, and the fingers all point in the direction of the magnetic field, then the palm will push in the direction of the force on the positive charge (proton).

From this, we can deduce from the original statement about this proton that the direction of the field is into the screen of this computer. with that field direction held constant, we can work out that

a) if the thumb point down, the force will be to the right

b) if the thumb points to the right, the force will be upwards

c)  if the thumb points to the left, the force is downwards

d)  if the thumb points into the page, then there will be no force on the proton since the proton must travel perpendicularly to the magnetic field for a force to be induced on it

e) explanation is the same as foe option d.

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Answer:

The magnitude of the acceleration  is a = 0.33 m/s^2

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            = 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \  X \ \  \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \  \r  j))

            = (-5.966*10^4 N) \r k

Note :

           i \ \ X \ \ j = k \\\\j \ \  X  \ \ k = i\\\\k  \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \  X  \ \ j = -i\\\\i  \ \ X \ \ k = - j\\

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Making a the subject

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   Substituting values

     a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}

        = (-0.33m/s^2)\r k

        = 0.33m/s^2 * (- \r k)

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