A proton accelerated through a potential difference of 3.0kV enters a magnetic field of flux density 0.1T, in a plane perpendicular to the motion of the proton. Calculate?

The force experienced by a charged particle moving through a magnetic field is given by the equation:

F = qvB

Where:
F = force
q = charge of the particle (in this case, the charge of a proton is 1.6 x 10^-19 C)
v = velocity of the particle
B = magnetic flux density

The velocity of the proton can be calculated using the equation for kinetic energy:

KE = qV

Where:
KE = kinetic energy
q = charge of the particle
V = potential difference

Rearranging the equation for velocity:

v = sqrt(2KE/m)

Where:
m = mass of the proton (1.67 x 10^-27 kg)

Substitute the values into the equations:

KE = 1.6 x 10^-19 C * 3.0 x 10^3 V = 4.8 x 10^-16 J

v = sqrt(2 * 4.8 x 10^-16 J / 1.67 x 10^-27 kg) = 1.6 x 10^7 m/s

Now, calculate the force experienced by the proton:

F = 1.6 x 10^-19 C * 1.6 x 10^7 m/s * 0.1 T = 1.6 x 10^-12 N

Therefore, the force experienced by the proton is 1.6 x 10^-12 N.

1. Proton's kinetic energy:

Kinetic energy = qV = (1.6 x 10^-19 C)(3000 V) = 4.8 x 10^-16 J

2. Proton's velocity:

Kinetic energy = 1/2 - mv^2
4.8 x 10^-16 J = 1/2 - (1.67 x 10^-27 kg) - v^2
v = 1.2 x 10^6 m/s

3. Magnetic force on proton:

Magnetic force = qvB
Magnetic force = (1.6 x 10^-19 C)(1.2 x 10^6 m/s)(0.1 T)
Magnetic force = 1.92 x 10^-13 N

4. Radius of circular path:

Magnetic force = mv^2/r
1.92 x 10^-13 N = (1.67 x 10^-27 kg)(1.2 x 10^6 m/s)^2 / r
r = 0.01 m = 1 cm

Therefore, the proton will move in a circular path with a radius of 1 cm.