13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path o
1 answer:
The magnetic force acting on a charged particle moving perpendicular to the field is:
= qvB
is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.
The centripetal force acting on a particle moving in a circular path is:
= mv²/r
is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.
If the magnetic force is acting as the centripetal force, set equal to
and solve for B:
qvB = mv²/r
B = mv/(qr)
Given values:
m = 1.67×10⁻²⁷kg (proton mass)
v = 7.50×10⁷m/s
q = 1.60×10⁻¹⁹C (proton charge)
r = 0.800m
Plug these values in and solve for B:
B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)
B = 0.979T
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Answer:
Explanation:
Given:
height above which the rock is thrown up,
initial velocity of projection,
let the gravity on the other planet be g'
The time taken by the rock to reach the top height on the exoplanet:
where:
final velocity at the top height = 0
(-ve sign to indicate that acceleration acts opposite to the velocity)
The time taken by the rock to reach the top height on the earth:
Height reached by the rock above the point of throwing on the exoplanet:
where:
final velocity at the top height = 0
Height reached by the rock above the point of throwing on the earth:
The time taken by the rock to fall from the highest point to the ground on the exoplanet:
(during falling it falls below the cliff)
here:
initial velocity= 0
Similarly on earth:
Now the required time difference: