If a charged particle of charge to mass ratio $(q / m) = α$ enters in a magnetic field of strength $B$ at a speed $v = (2 α d) (B)$ , then :

Question

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Answer 1

Radius of the portion of the circular track inside magnetic field: $r = \frac{m v}{q B} = \frac{v}{\frac{q}{m} B} = \frac{2 α d B}{α B} = 2 d$
Time period $T = \frac{2 π m}{q B} = \frac{2 π}{\frac{q}{m} B} = \frac{2 π}{α B}$
The particle spends half of $T$ inside magnetic field.
Time spent inside magnetic field is $\frac{T}{2} = \frac{1}{2} \frac{2 π}{α B} = \frac{π}{α B}$
Angle subtended by the path of charged particle in magnetic field at the center of circular path is $π$ .

Answer 2

Radius of the portion of the circular track inside magnetic field:

r = \frac{m v}{q B} = \frac{v}{\frac{q}{m} B} = \frac{2 α d B}{α B} = 2 d

Time period

T = \frac{2 π m}{q B} = \frac{2 π}{\frac{q}{m} B} = \frac{2 π}{α B}

The particle spends half of

T

inside magnetic field.
Time spent inside magnetic field is

\frac{T}{2} = \frac{1}{2} \frac{2 π}{α B} = \frac{π}{α B}

Angle subtended by the path of charged particle in magnetic field at the center of circular path is

π

.