One common type of cosmic ray is a proton traveling at close to the speed of light. If the proton is traveling downward, as shown in the figure below, at a speed of 2.03 ✕ 10^7 m/s, what are the magnitude and direction of the magnetic field at point B?
www.webassign.net/katzpse1/30-p-014.png
To determine the magnitude and direction of the magnetic field at point B, we can use the right-hand rule for the magnetic force on a moving charge.
The right-hand rule states that if you point your right thumb in the direction of the moving charge (proton), and curl your fingers towards the direction of the magnetic field, your palm will face the direction of the magnetic force.
In this case, the proton is traveling downward, so we point our thumb downward. Now, we need to curl our fingers towards the direction of the magnetic field at point B. Looking at the figure in the link you provided, the magnetic field lines appear to be going into the page in a clockwise direction. Therefore, we curl our fingers in a clockwise direction.
By curling our fingers downward and in a clockwise direction, our palm is facing towards the right. This indicates that the magnetic force on the proton is directed towards the right.
To find the magnitude of the magnetic field at point B, we can use the equation for the magnetic force on a moving charge. The magnetic force (F) is given by the equation F = qvB, where q is the charge of the particle (proton), v is the velocity of the particle, and B is the magnetic field strength.
In this case, we are given the velocity of the proton as 2.03 x 10^7 m/s. The charge of a proton is 1.6 x 10^-19 C. Hence, the magnetic force on the proton is equal to qvB.
Since the proton is traveling in a straight line (not being deflected by the magnetic field), the magnetic force is equal to the downward gravitational force on the proton. Therefore, we have F = mg, where m is the mass of the proton and g is the acceleration due to gravity.
By equating mg to qvB, we can solve for B:
qvB = mg
B = (mg) / (qv)
Now, we need to calculate the mass of the proton and the charge of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg, and the charge of a proton is 1.6 x 10^-19 C.
Plugging in the values, we get:
B = (1.67 x 10^-27 kg × 9.8 m/s^2) / (1.6 x 10^-19 C × 2.03 x 10^7 m/s)
Evaluating this expression, we find:
B ≈ 1.1 x 10^-5 T
Therefore, the magnitude of the magnetic field at point B is approximately 1.1 x 10^-5 Tesla, and the direction of the magnetic field is towards the right.