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An electron in a hydrogen atom revolves around its nucleus with a speed of   6.76\times10^{6}ms^{-1}   in an orbit of radius 0.52 A^{0}  . The magnetic field produced at the nucleus of the hydrogen atom is ______ T.

Option: 1

40


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Magnetic field due to moving charge
 
\begin{aligned} & \mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2} \\ & B=\frac{\mu_0}{4 \pi} \frac{\mathrm{ev} \sin (\pi / 2)}{r^2} \end{aligned}

 \begin{aligned} & \mathrm{~B}=10^{-7} \times \frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{\left(0.52 \times 10^{-10}\right)^2} \\ & \mathrm{~B}=40 \mathrm{~T} \end{aligned}

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himanshu.meshram

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