Question

На заряд 2 мкКл, що рухається в однорідному магнітному полі зі швидкістю 5·10 6 м/с під кутом 45º до ліній індукції магнітного поля, діє сила 2,8мН. Визначте модуль вектора магнітної індукції.

Answer 1

Ответ:

відповідь: 0,4*10'-3

Answer 2

Calculation of Magnetic Induction

To determine the magnitude of the magnetic induction vector, we can use the formula for the magnetic force on a moving charged particle in a magnetic field:

F = q * v * B * sin(θ)

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

In this case, we are given the following values: - Charge of the particle (q) = 2 μC = 2 * 10^-6 C - Velocity of the particle (v) = 5 * 10^6 m/s - Angle between the velocity and the magnetic field (θ) = 45º - Magnetic force (F) = 2.8 mN = 2.8 * 10^-3 N

We can rearrange the formula to solve for the magnetic field induction (B):

B = F / (q * v * sin(θ))

Let's calculate the value of B using the given values:

B = (2.8 * 10^-3 N) / (2 * 10^-6 C * 5 * 10^6 m/s * sin(45º))

Using a calculator, we can simplify the equation further:

B = (2.8 * 10^-3 N) / (10^-5 C * 5 * 10^6 m/s * 0.7071)

B = (2.8 * 10^-3 N) / (3.535 * 10^-5 C * m/s)

B ≈ 79.07 T

Therefore, the magnitude of the magnetic induction vector is approximately 79.07 T.