Question

Под каким углом к линиям индукции движется протон в магнитном поле с индукцией 0,2 Тл, если сила действующая на протон равна 20 фН при скорости движения 5Мм/с?

Answer 1

F=v*b*q*(sinA)
sinA= f/v*b*q

Answer 2

Calculating the Angle of Deflection for a Proton in a Magnetic Field

To calculate the angle at which a proton moves in a magnetic field with an induction of 0.2 T, given a force of 20 fN and a velocity of 5 Mm/s, we can use the formula for the force on a charged particle moving in a magnetic field:

F = qvBsin(θ)

Where: - F is the force on the particle - 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

First, we need to solve for the angle θ using the given values.

Calculating the Angle

Using the given force, velocity, and magnetic field induction, we can rearrange the formula to solve for the angle θ:

θ = sin^(-1)(F / (qvB))

Substitute the given values: - F = 20 fN = 20 x 10^(-15) N - q = charge of a proton = 1.6 x 10^(-19) C - v = 5 Mm/s = 5 x 10^6 m/s - B = 0.2 T

Now, let's calculate the angle θ using the given values.

Calculation

Using the formula θ = sin^(-1)(F / (qvB)): ``` θ = sin^(-1)((20 x 10^(-15)) / (1.6 x 10^(-19) x 5 x 10^6 x 0.2)) ``` Calculating the angle using the given values: ``` θ ≈ sin^(-1)(0.025 / (1.6 x 5 x 0.2)) ≈ sin^(-1)(0.025 / 1.6) ≈ sin^(-1)(0.015625) ≈ 0.946 degrees ```

Conclusion

Therefore, the angle at which the proton moves in the magnetic field with an induction of 0.2 T, given a force of 20 fN and a velocity of 5 Mm/s, is approximately 0.946 degrees.