Electron moves at right angles to a magnetic | vedclass

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Question

$B=1.5 	imes 10^{-2} \,\mathrm{T}$

$	heta=90^{\circ}, \sin 	heta=1, \mathrm{v}=6 	imes 10^{7} \,\mathrm{m} / \mathrm{s}$

$\frac{\mathrm{e}}{

Vedclass · Accepted Answer

$2.35$

Vedclass · Answer

$B=1.5 	imes 10^{-2} \,\mathrm{T}$

$	heta=90^{\circ}, \sin 	heta=1, \mathrm{v}=6 	imes 10^{7} \,\mathrm{m} / \mathrm{s}$

$\frac{\mathrm{e}}{\mathrm{m}}=1.7 	imes 10^{11} \,\mathrm{C} / \mathrm{kg}$

$\mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}=\frac{6 	imes 10^{7}}{1.5 	imes 10^{-2} 	imes 1.7 	imes 10^{11}}$

$=2.35 	imes 10^{-2}\, \mathrm{m}=2.35\, \mathrm{cm}$

Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}\,tesla$ with speed of $6 \times 10^7\,m/s$. If the specific charge of the electron is $1.7 \times 10^{11}\,C/kg$. The radius of circular path will be......$cm$

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