Question

Протон, ускоренный разностью потенциалов 1,5 кВ, влетает в однородное магнитное поле с индукцией 20 мТл. Скорость протона перпендикулярна направлению вектора магнитной индукции. Определите радиус окружности, по которой движется протон, и период вращения протона.

Answer 1

Вывод формулы на картинке

Answer 2

Calculation of the Radius and Period of the Proton's Circular Motion

To determine the radius and period of the proton's circular motion, we can use the formula for the centripetal force acting on a charged particle moving in a magnetic field.

The centripetal force is given by the equation:

F = qvB

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

In this case, the proton is accelerated by a potential difference of 1.5 kV, which means it gains kinetic energy. We can use the kinetic energy formula to find the velocity of the proton:

K.E. = (1/2)mv^2

Where: - K.E. is the kinetic energy, - m is the mass of the proton (approximately 1.67 x 10^-27 kg), - v is the velocity of the proton.

By equating the kinetic energy gained by the proton to the potential energy gained from the potential difference, we can find the velocity of the proton.

Now let's calculate the radius and period of the proton's circular motion.

Calculation Steps:

1. Calculate the velocity of the proton using the kinetic energy formula:

K.E. = (1/2)mv^2

Since the proton gains kinetic energy from the potential difference, we can equate the kinetic energy gained to the potential energy gained:

K.E. = qV

Where: - q is the charge of the proton (1.6 x 10^-19 C), - V is the potential difference (1.5 kV).

Equating the two equations:

(1/2)mv^2 = qV

Solving for v:

v = sqrt((2qV)/m)

Substituting the values:

v = sqrt((2 * 1.6 x 10^-19 C * 1500 V) / (1.67 x 10^-27 kg))

Calculating the value of v using the given values:

v ≈ 1.51 x 10^7 m/s

2. Calculate the radius of the proton's circular motion using the formula for the centripetal force:

F = qvB

The centripetal force is provided by the magnetic field, so we can equate the centripetal force to the magnetic force:

qvB = mv^2 / r

Solving for r:

r = mv / (qB)

Substituting the values:

r = (1.67 x 10^-27 kg * 1.51 x 10^7 m/s) / (1.6 x 10^-19 C * 20 x 10^-3 T)

Calculating the value of r using the given values:

r ≈ 0.124 meters

3. Calculate the period of the proton's circular motion using the formula:

T = (2πr) / v

Substituting the values:

T = (2π * 0.124 meters) / (1.51 x 10^7 m/s)

Calculating the value of T using the given values:

T ≈ 2.61 x 10^-8 seconds

Therefore, the radius of the circular path along which the proton moves is approximately 0.124 meters, and the period of the proton's rotation is approximately 2.61 x 10^-8 seconds.

Please note that the calculations provided are based on the given information and assumptions made.