Question

Протон , влетев в магнитное поле с индукцией 0,02 Тл , описал окружность радиусом 10 см. Найти его скорость.

Answer 1

B=0.02

q=1.6*10^-19

m=1.67*10^-27

R=0.1

q*B*v=m*v²/R

v=q*B*R/m=.........

Answer 2

Calculation of the Proton's Velocity

To find the velocity of the proton, we can use the formula for the centripetal force acting on a charged particle moving in a magnetic field:

F = qvB,

where: - F is the centripetal force, - q is the charge of the particle, - v is the velocity of the particle, and - B is the magnetic field strength.

In this case, the proton is moving in a magnetic field with an induction of 0.02 T and is following a circular path with a radius of 10 cm.

Since the centripetal force is provided by the magnetic force, we can equate the two:

qvB = mv^2 / r,

where: - m is the mass of the proton, and - r is the radius of the circular path.

We can rearrange the equation to solve for the velocity v:

v = (qBr / m)^0.5.

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

Using the charge of a proton q = 1.6 x 10^-19 C and the mass of a proton m = 1.67 x 10^-27 kg, and converting the radius to meters r = 0.1 m, and the magnetic field strength B = 0.02 T, we can calculate the velocity.

v = ((1.6 x 10^-19 C) x (0.02 T) x (0.1 m)) / (1.67 x 10^-27 kg))^0.5.

Calculating the above expression gives us the velocity of the proton.

Please note that the above calculation assumes that the proton is moving perpendicular to the magnetic field.

Let me calculate the velocity for you.