Question

a proton is moving along y-axis with velocity 200 m/s and mag. field 1 micro tesla acting at an angle 30 degree to y-axis;radius of its helical path will be

Answer 1

Answer:

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Answer 2

Given:

The velocity of the proton along y-axis = 200m/s

The magnitude of magnetic field = 1μT = 10⁻⁶T

The angle from the y-axis = 30°

To Find:

The radius of the helical path

Solution:

A charged particle follows a helical path when it enters a uniform magnetic field B at an angle θ.

The centripetal force on a charged particle in uniform B = $F_c=\frac{mv^2}{r}$

Applying Newton's second law of motion (F = ma) and balancing the centripetal force with the magnetic force we get:

q B v sinθ = $\frac{mv^2sin^2\theta}{r}$

or R = $\frac{mvsin\theta}{qB}$

The charge q on a proton is 1.6 X 10⁻¹⁹C

and mass of proton = 1.6 X 10⁻²⁷ kg

Substituting,

→ R = 1.6 X 10⁻²⁷ X 200 X 0.5 / 1.6 X 10⁻¹⁹ X 10⁻⁶

= 100 X 10⁻²⁷ / 10⁻¹⁹ X 10⁻⁶

= 1 m

Hence, the radius of the proton's helical path will be 1m.