The magnetic field due to the electron is given by
B=4πr2μ0qvsin90∘
The given data is
v=6.76×106ms−1r=0.52×10−10m
Thus, the magnetic field is
B=0.52×0.52×10−2010−7×(1.6××10−19)×(6.76×106)=40T
JEE Main 2023 — Physics Electromagnetism
An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×106ms–1 in an orbit of radius 0.52A˚. The magnetic field produced at the nucleus of the hydrogen atom is _______ T.
Held on 15 Apr 2023 · Verified 6 Jul 2026.
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A circular loop of radius $20\text{ cm}$ and resistance $2\ \Omega$ is placed in a time varying magnetic field $\vec{B} = (2t^2 + 2t + 3)\text{ T}$. At $t = 0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t = 3\text{ s}$ is $\dfrac{\alpha}{50}\text{ A}$. The value of $\alpha$ is _______. (Take $\pi = 22/7$)
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