Question

Протон разгоняется разностью потенциалов 1,5 кВ и попадает в однородное магнитное поле перпендикулярно его силовым линиям. Радиус кривизны окружности, по которой протон движется в магнитном поле 56 см. Найти индукцию магнитного поля и указать его направление на рисунке.

Answer 1

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Answer 2

Calculation of Magnetic Field Induction

To find the magnetic field induction (B) and its direction, we can use the formula for the centripetal force acting on a charged particle moving in a magnetic field:

F = (mv^2) / r

Where: - F is the centripetal force - m is the mass of the particle - v is the velocity of the particle - r is the radius of the circular path

In this case, the centripetal force is provided by the electric field, which is given by the potential difference (V) between the proton's initial and final positions:

F = qV

Where: - q is the charge of the proton

Since the proton is accelerated by a potential difference of 1.5 kV, we can equate the two formulas:

(mv^2) / r = qV

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

B = (mv) / (q * r)

Now, let's substitute the given values into the equation:

- Radius of the circular path (r) = 56 cm = 0.56 m - Potential difference (V) = 1.5 kV = 1500 V - Charge of a proton (q) = 1.6 x 10^-19 C (elementary charge) - Mass of a proton (m) = 1.67 x 10^-27 kg

B = (1.67 x 10^-27 kg * v) / (1.6 x 10^-19 C * 0.56 m)

Unfortunately, the velocity (v) of the proton is not provided in the question. Without the velocity, we cannot calculate the magnetic field induction (B). Please provide the velocity of the proton so that we can proceed with the calculation.