An α -particle with a specific charge of 2.5 × 10 7 C k g - 1 moves with a speed of 2 × 10 5 m s - 1 in a perpendicular magnetic field of 0.05 T . Then, the radius of the circular path described by it is

r = m v B q ⇒ r = v B q m = 2 × 10 5 0.05 × 2.5 × 10 7 = 2 × 10 7 12.5 × 10 7 = 200 12.5 c m = 16 c m

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic ForceEasy2 minPYQ_2020

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An $α$ -particle with a specific charge of $2.5 \times 10^{7} C {k g}^{- 1}$ moves with a speed of $2 \times 10^{5} m s^{- 1}$ in a perpendicular magnetic field of $0.05 T$ . Then, the radius of the circular path described by it is

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Answer:

Solution:

$r = \frac{m v}{B q}$

$\Rightarrow r = \frac{v}{B \frac{q}{m}} = \frac{2 \times 10^{5}}{0.05 \times 2.5 \times 10^{7}}$

$= \frac{2 \times 10^{7}}{12.5 \times 10^{7}} = \frac{200}{12.5} c m = 16 c m$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force

⏱ 2mℹ️ Source: PYQ_2020

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