The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
Answer: C. T
Explanation:
Period is a unit of time; and in the context of waves or oscillations it is defined as <em>"the </em><u><em>time </em></u><em>elapsed between two equivalent points on the wave or oscillation".</em>
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It is important to note Period (denoted by ) is one of the most important factors (along with the amplitude, frequency and velocity) to describe and characterize a wave.
In addition, Period has an inverse relation with the frequency , this means that if we are given the frequency of a wave, we can inmediatly know its Period.
Answer:
Magnitude of magnetic field is 1.29 x 10⁻⁴ T
Explanation:
Given :
Current flowing through the wire, I = 16.9 A
Length of wire. L = 0.69 m
Magnetic force experienced by the wire, F = 1.5 x 10⁻³ N
Consider B be the applied magnetic field.
The relation to determine the magnetic force experienced by current carrying wire is:
F = ILBsinθ
Here θ is the angle between magnetic field and current carrying wire.
According to the problem, the magnetic field and current carrying wire are perpendicular to each other, that means θ = 90⁰. So, the above equation becomes:
F = ILB
Substitute the suitable values in the above equation.
B = 1.29 x 10⁻⁴ T
Answer:
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr
Explanation:
Given data:
A position is 4km north and 2.5 km east to B
Ship A velocity = 22 km/hr
ship B velocity = 40 km/hr
A velocity wrt to velocity of B
putting respective value to get velocity of A with respect to B
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr