For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

$qvB = m\frac{v^2}{r}$

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

$T=\frac{2\pi r}{v}$

Re-arranging for r

$r=\frac{Tv}{2\pi}$

And substituting into the previous equation

$qvB = m \frac{Tv^3}{2\pi}$

Solving for T,

$T=\frac{2\pi q B}{m v^2}$

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0

3 years ago

Ask Your Teacher A basketball player shoots toward a basket 5.8 m away and 3.0 m above the floor. If the ball is released 1.7 m

const2013 [10]

Answer:

The answer to your question is vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

height 2 = 1.7 m

angle = 60°

vo = ?

g = 9.81 m/s²

Formula

hmax = vo²sinФ/ 2g

Solve for vo²

vo² = 2ghmax / sinФ

Substitution

vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

vo² = 19.62(1.3) / 0.866

vo² = 25.51 / 0.866

vo² = 29.45

Result

vo = 5.43 m/s

5 0

3 years ago