A 42 mT magnetic field points due west. If a proton of kinetic energy 9x10E-12 J enters this field in an upward direction, find the magnetic force acting on the proton in magnitude and direction.
What magnetic force would an electron experience due to Earth’s magnetic field if it were moving at 3x10E6 m/s directly away from earth’s surface? (Earth’s B-field in the region is parallel to the surface and has a strength of 550 µT)
A 60 cm long wire is oriented from east to west directly above the equator and carries a current of 5 A (toward the east). What is the magnitude and direction of the magnetic force on this wire due to the earth’s magnetic field (30 µT oriented south and north)?
To find the magnetic force acting on the proton in the first question, we can use the equation for the magnetic force on a charged particle in a magnetic field. The equation is given by:
F = q * v * B * sin(theta)
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, and
theta is the angle between the velocity vector and the magnetic field vector.
In this case, the charge of the proton (q) is equal to 1.6 x 10^-19 C, the velocity (v) is given as upward, so we can assume it's directly north, the magnetic field (B) is 42 mT pointing due west, and the angle between the velocity and magnetic field (theta) is 90 degrees since they are perpendicular to each other.
Now, we can substitute these values into the formula to calculate the force:
F = (1.6 x 10^-19 C) * (9 x 10^-12 J) * (42 x 10^-3 T) * sin(90 degrees)
Simplifying this equation, we get:
F = 6.048 x 10^-31 N pointing in the upward direction.
Therefore, the magnetic force acting on the proton is approximately 6.048 x 10^-31 N in magnitude, upwards.
Moving on to the second question, to find the magnetic force experienced by an electron due to Earth's magnetic field, we use the same formula:
F = q * v * B * sin(theta)
In this case, the charge of the electron is also 1.6 x 10^-19 C and the velocity (v) is given as directly away from the Earth's surface, so we can assume it's directly upwards. The magnetic field (B) is 550 µT, which points parallel to the Earth's surface, and the angle between the velocity and the magnetic field (theta) is 0 degrees since they're parallel.
Now, we substitute these values into the formula:
F = (1.6 x 10^-19 C) * (3 x 10^6 m/s) * (550 x 10^-6 T) * sin(0 degrees)
Since sin(0 degrees) = 0, the force acting on the electron is zero.
Therefore, the electron would not experience any magnetic force due to Earth's magnetic field.
In the last question, we can find the magnetic force on the wire using the formula for the magnetic force on a current-carrying wire in a magnetic field:
F = I * L * B * sin(theta)
Where:
F is the magnetic force,
I is the current,
L is the length of the wire,
B is the magnetic field, and
theta is the angle between the wire and the magnetic field.
In this case, the current (I) is 5 A, the length of the wire (L) is 60 cm (or 0.6 m), the magnetic field (B) is 30 µT, and the angle between the wire and the magnetic field (theta) is 90 degrees since the wire is perpendicular to the magnetic field.
Substituting these values into the formula, we get:
F = (5 A) * (0.6 m) * (30 x 10^-6 T) * sin(90 degrees)
Since sin(90 degrees) = 1, the force acting on the wire is:
F = (5 A) * (0.6 m) * (30 x 10^-6 T) * 1 = 9 x 10^-5 N
Therefore, the magnitude of the magnetic force acting on the wire is approximately 9 x 10^-5 N, directed south.