Question

calculate the speed of a proton, moving in a circular path of radius 8.0 cm, in a plane perpendicular to a uniform 1.5 t magnetic field. what voltage would be required to accelerate the proton from rest, in a vacuum, to this speed? (mp

Answer 1

To calculate the speed of a proton in a magnetic field, set the magnetic force equal to the centripetal force. Then, use the kinetic energy and electric potential energy relationship to find the voltage required to accelerate the proton from rest to the calculated speed.

Step-by-step explanation:

To calculate the speed of a proton moving in a circular path in a magnetic field, we can use the formula for the magnetic force, which is equal to the centripetal force required to keep the proton in circular motion:

• Fmagnetic = qvB
• Fcentripetal = (mv^2)/r

where:

• q is the charge of the proton (1.602 x 10^-19 C)
• v is the velocity of the proton
• B is the magnetic field strength (1.5 T)
• m is the mass of the proton (1.673 x 10^-27 kg)
• r is the radius of the circular path (0.08 m)

Setting the magnetic force equal to the centripetal force, we get:

qvB = (mv^2)/r

Solving for v, we find:

v = qBr / m

Substituting in the known values, we calculate the proton's speed.

As for the voltage required to accelerate the proton from rest to this speed in a vacuum, we use the kinetic energy and electric potential energy relationship:

• KE = qV
• (1/2)mv^2 = qV

Where V is the voltage needed. We then solve for V using the previously found value for v.