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4 years ago

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A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y

-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line.
The electric field should be in the:

a)neg y direction

b)neg x direction

c)postive x direction

d)postive z direction

e)negative z

1 answer:

Genrish500 [490]4 years ago

8 0

Answer:

Positive z direction.

Explanation:

The magnetic force acting on the electron is given by the formula as :

$F=q(v\times B)$

q is the charge on proton

v is the speed of proton

B is the magnetic field

It is mentioned that the proton is moving with a velocity in the positive x-direction. The uniform magnetic field B in the positive y-direction such taht,

q = +e

v = vi

B = Bj

$F=e(v\ i\times B\ j)$

Since, $i\times j=k$

$F=(evB)\ k$

So, the magnetic force acting on the proton in positive z axis. Hence, the correct option is (d) "positive z direction".

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The acceleration of gravity for an object near a planet is given by

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The critical velocity for a satellite orbiting around a planet is given by

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$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

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$R=6.37\cdot 10^6 m$

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v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

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where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

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