Question

a proton moves in a circle of radius 0.4 when it enters a region with a magnetic field of 1.0t which points into the plane the speed of the proton is

Answer 1

Answer:

4 x 10⁷m/s

Explanation:

When a charged particle moves in a curved path in a magnetic field, it experiences some magnetic force, and in the absence of any other force, which supplies the centripetal force needed to keep the particle in balance.

Let the magnetic force be $F_{M}$

Let the centripetal force be $F_{C}$

=> $F_{M}$ = $F_{C}$           --------------(i)

We know that;

$F_{M}$ = qvBsinθ

Where;

q = charge on the particle

v = speed of the particle

B = magnetic field intensity

θ = angle between the speed and magnetic field vectors

Also;

$F_{C}$ = $\frac{mv^2}{r}$

Where;

m = mass of the particle

v = velocity/speed of the particle

r = radius of the circular path of motion.

From equation (i)

qvBsinθ = $\frac{mv^2}{r}$           [divide both sides by v]

qBsinθ = $\frac{mv}{r}$              [make v subject of the formula]

v = qrBsinθ / m            --------------------(ii)

From the question;

B = 1.0T

r = 0.4m

θ = 90°     [since magnetic field is always perpendicular to velocity]

q = 1.6 x 10⁻¹⁹C        [charge of a proton]

m = 1.6 x 10⁻²⁷kg        [mass of a proton]

Substitute these values into equation(ii) as follows;

v = (1.6 x 10⁻¹⁹ x 0.4 x 1.0 x sin90°) /  (1.6 x 10⁻²⁷)

v = 4 x 10⁷ m/s

Therefore the speed of the proton is 4 x 10⁷m/s