A proton moving with a velocity 3 × 1 0 5 m / s enters a magnetic field of 0.3 tesla at an angle of 3 0 ∘ with the field. The radius of curvature of its path will be ( e / m for proton = 1 0 8 C / kg )

r = B e m v s i n θ = 0.3 × 1 0 8 3 × 1 0 5 s i n 3 0 ∘ 3 × 1 0 7 3 × 1 0 5 × 2 1 = 0.5 × 1 0 − 2 m = 0.5 cm

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PhysicsMagnetismMagnetic ForceMedium2 minPYQ_2021

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A proton moving with a velocity $3 \times 1 0^{5} m / s$ enters a magnetic field of $0.3$ tesla at an angle of $3 0^{\circ}$ with the field. The radius of curvature of its path will be $(e / m$ for proton $= 1 0^{8} C / kg$ )

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Answer:

Solution:

$r = \frac{m v s i n θ}{B e} = \frac{3 \times 1 0 ^{5} s i n 3 0 ^{\circ}}{0.3 \times 1 0 ^{8}}$ $\frac{3 \times 1 0 ^{5} \times \frac{1}{2}}{3 \times 1 0 ^{7}} = 0.5 \times 1 0^{- 2}$ $m = 0.5 cm$

Stream:BITSATSubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force

⏱ 2mℹ️ Source: PYQ_2021

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