If a proton is moving in a magnetic field with a speed of 4 × 106m/s and is executing uniform circular motion with a radius of 10 cm, what is the magnetic field?
To find the magnetic field, we can use the equation for the centripetal force of a charged particle moving in a magnetic field.
The centripetal force (F) is given by the equation:
F = (mv^2)/r
Where:
m = mass of the proton (1.67 x 10^-27 kg)
v = velocity of the proton (4 x 10^6 m/s)
r = radius of the circular path (0.10 m)
The centripetal force is provided by the magnetic force (Fm) acting on the proton. The magnetic force can be calculated using the equation:
Fm = qvB
Where:
q = the charge on the proton (+1.6 x 10^-19 C)
v = velocity of the proton (4 x 10^6 m/s)
B = magnetic field strength (unknown)
Since the centripetal force is equal to the magnetic force, we can equate the two equations:
qvB = (mv^2)/r
Now we can solve for the magnetic field (B):
B = (mv)/qr
B = (1.67 x 10^-27 kg)(4 x 10^6 m/s) / ((+1.6 x 10^-19 C)(0.10 m))
Evaluating this expression gives the value of the magnetic field.