Imaging that an electron revolves around a circle of the radius 5 .3 × 10 - 11 m with a linear speed of 7.5 × 10 4 m s - 1 in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is

B = μ 0 4 π e v r 2 = 10 - 7 × 1.6 × 10 - 19 × 7.5 × 10 + 4 5.3 × 10 - 11 2 On solving, B = 0.43 W b m - 2

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismBiot-savart's & Ampere's Circuital lawEasy2 minPYQ_2020

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Imaging that an electron revolves around a circle of the radius $5 .3 \times 10^{- 11} m$ with a linear speed of $7.5 \times 10^{4} m s^{- 1}$ in a hydrogen atom. The magnetic field produced at the centre of the circle, due to the electron, is

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Answer:

Solution:

$B = \frac{μ_{0}}{4 π} \frac{e v}{r^{2}}$

$= \frac{10^{- 7} \times 1.6 \times 10^{- 19} \times 7.5 \times 10^{+ 4}}{{(5.3 \times 10^{- 11})}^{2}}$

On solving, $B = 0.43 W b m^{- 2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:MagnetismSubtopic:Biot-savart's & Ampere's Circuital law

⏱ 2mℹ️ Source: PYQ_2020

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