The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 54.0 µT. A proton is moving horizontally towards the west in this field with a speed of 6.00 ✕ 106 m/s.
(a) What is the direction and magnitude of the magnetic force the field exerts on this charge?
N
(b) What is the radius of the circular arc followed by this proton?
km
F=qvBsinα
sinα=1
q=e=1.6•10⁻¹⁹ C
F=evB=.....
B↓v←F⊙
mv²/R=evB
R=mv/eB=.....
To find the direction and magnitude of the magnetic force exerted on the proton, we can use the formula for the magnetic force experienced by a charged particle moving in a magnetic field:
F = qvB sin(theta)
where:
F is the magnetic force vector,
q is the charge of the particle (in this case the charge of a proton, which is 1.6 x 10^-19 C),
v is the velocity vector of the particle (6.00 x 10^6 m/s towards the west),
B is the magnetic field vector (54.0 µT vertically downward), and
theta is the angle between the velocity vector and the magnetic field vector.
(a) To find the direction and magnitude of the force, we need to determine the angle theta. Since the velocity vector is towards the west and the magnetic field vector is vertically downward, the angle between them is 90 degrees, or pi/2 in radians.
Using the formula for the magnetic force, we can now calculate the magnitude of the force:
F = (1.6 x 10^-19 C) * (6.00 x 10^6 m/s) * (54.0 x 10^-6 T) * sin(pi/2)
= 5.44 x 10^-13 N
Therefore, the magnitude of the magnetic force exerted on the proton is approximately 5.44 x 10^-13 N.
(b) To find the radius of the circular arc followed by the proton, we can use the equation for the centripetal force experienced by an object moving in a circular path:
F = mv^2 / r
where:
F is the force,
m is the mass of the proton (approximately 1.67 x 10^-27 kg),
v is the velocity of the particle (6.00 x 10^6 m/s towards the west), and
r is the radius of the circular path.
Since the magnetic force is providing the centripetal force, we can equate the forces:
F = mv^2 / r
5.44 x 10^-13 N = (1.67 x 10^-27 kg) * (6.00 x 10^6 m/s)^2 / r
Rearranging the equation to solve for the radius:
r = (1.67 x 10^-27 kg) * (6.00 x 10^6 m/s)^2 / 5.44 x 10^-13 N
Calculating the value:
r = 1.67 x 10^-27 kg * (6.00 x 10^6 m/s)^2 / 5.44 x 10^-13 N
≈ 1.05 x 10^-2 m
Therefore, the radius of the circular arc followed by the proton is approximately 1.05 x 10^-2 meters, or 10.5 mm.
To find the direction and magnitude of the magnetic force exerted on the proton, we can use the formula for the magnetic force on a moving charged particle:
F = q * v * B * sin(θ)
Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field,
θ is the angle between the velocity vector and the magnetic field vector.
In this case, we are given that the proton is moving horizontally towards the west, which means that the angle between its velocity vector and the vertical downward magnetic field is 90 degrees.
(a) Direction of the Magnetic Force:
Since sin(90 degrees) is equal to 1, we can ignore this term in our calculations. Therefore, the direction of the magnetic force is determined by the right-hand rule. If the thumb of your right hand points in the direction of the velocity vector (towards the west) and your fingers point in the direction of the magnetic field vector (vertically downward), then the palm of your hand will point in the direction of the magnetic force vector, which will be towards the south.
(b) Magnitude of the Magnetic Force:
To find the magnitude of the magnetic force, we need the charge of the proton. The charge of a proton is approximately equal to +1.602 × 10^-19 coulombs. Also, the magnitude of the magnetic field is given as 54.0 µT, which is equal to 54.0 × 10^-6 T. Plugging these values into the formula, we get:
F = (1.602 × 10^-19 C) * (6.00 × 10^6 m/s) * (54.0 × 10^-6 T)
Calculating this expression, we find that the magnitude of the magnetic force is approximately 5.19 × 10^-15 N.
Therefore, the answers are:
(a) The direction of the magnetic force is towards the south.
(b) The magnitude of the magnetic force is approximately 5.19 × 10^-15 N.
To find the radius of the circular arc followed by the proton, we can use the following formula relating the magnetic force, velocity, and radius:
F = (mv^2) / r
Where:
F is the magnetic force,
m is the mass of the proton,
v is the velocity of the proton,
r is the radius of the circular arc.
By rearranging the formula, we can solve for the radius:
r = (mv) / F
We already have the values for the mass of the proton (approximately 1.67 × 10^-27 kg), the velocity (6.00 × 10^6 m/s), and the magnetic force (5.19 × 10^-15 N). Plugging in these values, we can calculate the radius:
r = (1.67 × 10^-27 kg) * (6.00 × 10^6 m/s) / (5.19 × 10^-15 N)
Calculating this expression, we find that the radius of the circular arc followed by the proton is approximately 1.92 × 10^-2 meters, which is equal to 19.2 mm. Converting this to kilometers, we get:
Radius = 1.92 × 10^-2 m = 1.92 × 10^-5 km
Therefore, the answer is:
(b) The radius of the circular arc followed by the proton is approximately 1.92 × 10^-5 km.