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The proton moves through a magnetic field of magnitude 3x10^-5 T with a velocity of 4x10^5 m/s. Find the radius of the circular path the proton follows and indicate its direction of rotation.
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set the centripetal force equal to magnetic force, and solve for r.
To find the radius of the circular path the proton follows in the magnetic field, we can use the equation for the radius of the circular motion of a charged particle in a magnetic field:
r = (mv) / (qB)
Where:
r is the radius of the circular path,
m is the mass of the proton (= 1.67 × 10^-27 kg),
v is the velocity of the proton (= 4 × 10^5 m/s),
q is the charge of the proton (= 1.6 × 10^-19 C), and
B is the magnetic field strength (= 3 × 10^-5 T).
Now, let's plug in the given values into the equation:
r = [(1.67 × 10^-27 kg) × (4 × 10^5 m/s)] / [(1.6 × 10^-19 C) × (3 × 10^-5 T)]
Simplifying the expression:
r = (6.68 × 10^-22 kg·m/s) / (4.8 × 10^-24 C·T)
Dividing these values:
r ≈ 1.4 × 10^-2 m
So, the radius of the circular path the proton follows is approximately 1.4 × 10^-2 meters.
Now, to determine the direction of rotation, we can use the right-hand rule. If we point the right thumb in the direction of the proton's velocity (in this case, from the user-given values, we will point the thumb up), and curling the fingers towards the magnetic field direction (from the user-given values, we will curl the fingers towards the right), then the direction in which the fingers point gives the direction of rotation.
Therefore, using the right-hand rule, the proton would rotate in a clockwise direction.