Question

Протон движется со скоростью 10 см/с перпендикулярно однородному магнитному полю с индукцией 1 Тл. Найти радиус окружности, по которой он движется

Answer 1

Сила, действующая на протон, - это сила Лоренца. Угол 90°, значит, sinA=1:

Fл=BuqsinA=1*10*1,6*10⁻¹⁹*1=16*10⁻¹⁹Н

Т.к. протон движется по окружности, то Fл=Fцс:

Buq=mu²/R

R=mu/Bq=1,6*10⁻²⁷*10/(1*10⁻¹⁹)=16*10⁻⁸м=160нм

Answer 2

Calculation of the Radius of the Circular Path

To find the radius of the circular path along which the proton is moving, we can use the formula for the centripetal force acting on a charged particle in a magnetic field.

The centripetal force is given by the equation:

F = q * v * B

Where: - F is the centripetal force - q is the charge of the particle - v is the velocity of the particle - B is the magnetic field strength

In this case, the charge of the proton (q) is equal to the elementary charge (e), which is approximately 1.6 x 10^-19 Coulombs.

The velocity of the proton (v) is given as 10 cm/s.

The magnetic field strength (B) is given as 1 Tesla (T).

Substituting the values into the equation, we have:

F = (1.6 x 10^-19 C) * (10 cm/s) * (1 T)

To find the radius of the circular path, we can equate the centripetal force to the magnetic force acting on the proton:

F = q * v * B = m * v^2 / r

Where: - m is the mass of the proton - r is the radius of the circular path

The mass of the proton (m) is approximately 1.67 x 10^-27 kg.

Substituting the values into the equation, we have:

(1.6 x 10^-19 C) * (10 cm/s) * (1 T) = (1.67 x 10^-27 kg) * (10 cm/s)^2 / r

Simplifying the equation, we can solve for the radius (r):

r = (1.67 x 10^-27 kg) * (10 cm/s)^2 / [(1.6 x 10^-19 C) * (1 T)]

Now, let's calculate the value of the radius using the given values:

r = (1.67 x 10^-27 kg) * (10 cm/s)^2 / [(1.6 x 10^-19 C) * (1 T)]

r = (1.67 x 10^-27 kg) * (10^2 m^2/s^2) / [(1.6 x 10^-19 C) * (1 T)]

r = (1.67 x 10^-27 kg) * (10^2 m^2/s^2) / (1.6 x 10^-19 C)

r = (1.67 x 10^-27 kg) * (10^2 m^2/s^2) / (1.6 x 10^-19 C)

r ≈ 1.04 x 10^-5 meters or 10.4 micrometers

Therefore, the radius of the circular path along which the proton is moving is approximately 1.04 x 10^-5 meters or 10.4 micrometers.