Question

Решите задачу, пожалуйста. Желательно дано тоже написать Протон влетает в однородное магнитное поле с индукцией 250 тл под углом 60 градусов к силовым линиям. С какой скоростью движется протон, если на него действует сила 0,2 мкН.

Answer 1

Ответ:

Объяснение:

F = q·B·V·sin α

V = F / (q·B·sin α) = 0,2·10⁻⁶ / (1,6·10⁻¹⁹·250·10⁻³·0,866) ≈ 5,8·10⁶ м/с

Answer 2

Problem Statement:

A proton enters a uniform magnetic field with an induction of 250 T at an angle of 60 degrees to the field lines. If a force of 0.2 μN acts on the proton, what is its velocity?

Solution:

To solve this problem, we can use the formula for the magnetic force on a moving charged particle:

F = qvB sin(θ)

Where: - F is the magnetic force on the particle, - q is the charge of the particle, - v is the velocity of the particle, - B is the magnetic field induction, and - θ is the angle between the velocity vector and the magnetic field lines.

In this case, we are given: - F = 0.2 μN (0.2 x 10^-6 N), - B = 250 T, - θ = 60 degrees.

We need to find the velocity v of the proton.

To find the velocity, we can rearrange the formula as follows:

v = F / (qB sin(θ))

Now, let's substitute the given values into the formula and calculate the velocity.

v = (0.2 x 10^-6 N) / (1.6 x 10^-19 C * 250 T * sin(60 degrees))

Using the given values and performing the calculation, we find:

v ≈ 1.33 x 10^5 m/s

Therefore, the proton is moving with a velocity of approximately 1.33 x 10^5 m/s when a force of 0.2 μN acts on it in a magnetic field with an induction of 250 T at an angle of 60 degrees to the field lines.

Conclusion:

The proton is moving with a velocity of approximately 1.33 x 10^5 m/s when a force of 0.2 μN acts on it in a magnetic field with an induction of 250 T at an angle of 60 degrees to the field lines.