# Question #a8f63

Sep 11, 2015

$= 1 , 258 \mu m$

#### Explanation:

Note that the unit of magnetic flux density (B) is the Tesla, which is dimensionally equivalent to a $\frac{W b}{m} ^ 2$, since by definition it is the magnetic flux per unit area. (incorrect SI unit given in the original question).

The velocity v of the proton may be found from the definition of potential difference, together with the principle of conservation of energy. Since by definition, potential difference between 2 points is defined as the work done per unit charge to move a positive charge between the 2 points, and this work done is transferred into kinetic energy of the proton, it implies that
$Q V = \frac{1}{2} m {v}^{2} \implies v = \sqrt{\frac{2 Q V}{m}}$
$= \sqrt{\frac{\left(2\right) \left(9 , 1 \times {10}^{- 31}\right) \left(6 \times {10}^{5}\right)}{1 , 67 \times {10}^{- 27}}} = 25 , 571 m / s$

Now, an electric charge Q moving at velocity v in the presence of a magnetic field of flux density B, experiences a magnetic force given by
$\vec{F} = Q \vec{v} \times \vec{B}$

This is a vector cross product so its direction can be obtained from the right hand rule, and is perpendicular to both v and B.
This force is hence directed towards the centre all the time and is thus the required centripetal force ${F}_{c}$ to keep the charge moving in a circular motion.

The magnitude of this force can be given by $Q \cdot | | \vec{v} | | \cdot | | \vec{B} | | \cdot \sin \theta$ , where $\theta$ is the angle between $\vec{v} \mathmr{and} \vec{B}$

So, in this case, $Q v B \sin \theta = \frac{m {v}^{2}}{r}$
$\therefore r = \frac{m v}{Q B \sin \theta}$
$= \frac{\left(1 , 67 \times {10}^{- 27}\right) \left(25 , 571\right)}{\left(1 , 6 \times {10}^{-} 19\right) \left(0 , 3\right) \sin {45}^{\circ}}$

$= 1 , 258 \mu m$