Question

A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?

Answer 1

Answer:

0.17T

Explanation:

When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force $F_{M}$ which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force $F_{C}$ and the magnetic force. i.e

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

But;

$F_{C}$ = $\frac{mv^2}{r}$   [m = mass of the particle, r = radius of the path, v = velocity of the charge]

$F_{M}$ = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]

Substitute these into equation (i) as follows;

$\frac{mv^2}{r}$ = qvB

Make B subject of the formula;

B = $\frac{mV}{qr}$            ---------------(ii)

Known constants

m = 1.67 x 10⁻²⁷kg

q = 1.6 x 10⁻¹⁹C

From the question;

v = 1.4 x 10⁷m/s

r = 0.85m

Substitute these values into equation(ii) as follows;

B = $\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}$

B = 0.17T

Therefore, the magnetic field strength is 0.17T