An electron moves at right angle to a magnetic field of 5 × 1 0 − 2 T with a speed of 6 × 1 0 7 m / s . If the specific charge of the electron is 1.7 × 1 0 11 C / kg . The radius of the circular path will be

Radius of circular path r r = B mV = ( m e ) B V = 1.7 × 1 0 11 × 15 × 1 0 − 2 6 × 1 0 7 = 2.35 × 1 0 − 2 m = 2.35 m

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic ForceEasy2 minPYQ_2014

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An electron moves at right angle to a magnetic field of $5 \times 1 0^{- 2} T$ with a speed of $6 \times 1 0^{7} m / s$ . If the specific charge of the electron is $1.7 \times 1 0^{11} C / kg$ . The radius of the circular path will be

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Solution:

Radius of circular path $r r = \frac{mV}{B} = \frac{V}{( \frac{e}{m} ) B} = \frac{6 \times 1 0 ^{7}}{1.7 \times 1 0 ^{11} \times 15 \times 1 0 ^{- 2}} = 2.35 \times 1 0^{- 2} m = 2.35 m$

Stream:VITEEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic Force

⏱ 2mℹ️ Source: PYQ_2014

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