Question

A proton is projected with a velocity 10⁷ ms^-1 at right angles to a uniform magnetic field of induction 100 mT. The time (in seconds) taken by the proton to traverse 90^o arc is(Mass of proton = 1.65 × 10^-27 kg and charge of proton = 1.6 × 10^-19 C)

• (A) 0.81 × 10^-7
• (B) 1.62 × 10^-7
• (C) 2.43 × 10^-7
• (D) 3.24 × 10^-7

Answer 1

The Correct Option is B

Explanation:
When proton enters in a uniform magnetic field at right angle then it moves on a circular path. In this case velocity of proton.$$v=\frac{Bqr}{m}$$where $r=$ radius of circular path Distance covered by proton to traverse ${90}^{\circ }$Time taken by proton to cover distance $d$$$t=\frac{\pi r/2}{v}=\frac{\pi r}{2v}$$$$\begin{array}{l}t=\frac{\pi }{2}\frac{r}{Bqr/m}\\ t=\frac{\pi m}{2Bq}\end{array}$$Putting $m=1.65\times {10}^{-27}\mathrm{kg}$$$B=100\mathrm{mT}=100\times {10}^{-3}\mathrm{T},q=1.6\times {10}^{-19}\mathrm{C}$$$$\begin{array}{l}t=\frac{3.14\times 1.65\times {10}^{-27}}{2\times 100\times {10}^{-3}\times 1.6\times {10}^{-19}}\\ t=1.62\times {10}^{-7}\sec \end{array}$$

Answer 2

Consider a uniform magnetic field B acting perpendicular to the plane of paper and directed into the paper. Let a charged particle +q enter the region of the magnetic field with velocity v and perpendicular to the direction of B.

Force acting on the charge +q due to the uniform magnetic field is given by

F = qvB sinθ

Where θ is the angle between v and B.

Since the charged particle moves at a right angle to the magnetic field, therefore θ = 90⁰

⇒ F = qvB sin90 = qvB

This force always acts perpendicular to the direction of motion of the charged particle and magnetic field. Thus the path of the charged particle is Circular.

Now the necessary centripetal force required by the charged particle to move in a circular path is given by

F_C = mv²/r

Where

• m is the mass of the charged particle
• r is the radius of the circular path

This centripetal force is provided by the magnetic force F = qvB

Therefore, 

mv²/r = qvB

⇒ r = mv / qB

Time period to complete a circle is given by

T = circumference of the circle/speed

⇒ T = 2πr/v

On substituting the value of v, we get

T = 2πm/qB

Frequency of the charged particle is given by

f = 1/T = qB/2πm