A proton is moving with a uniform velocity of $2\times 1{0}^{8}\text{ }\text{m/s}$ in uniform magnetic and electric fields, which are perpendicular to each other. If the electric field is switched off, the proton moves in a circular path of radius $1.6\times 1{0}^{−5}\text{ }\text{m}$. The magnetic field ($B$) is:

(Shift II Memory Based)

Question

A proton is moving with a uniform velocity of $2\times 1{0}^{8}\text{ }\text{m/s}$ in uniform magnetic and electric fields, which are perpendicular to each other. If the electric field is switched off, the proton moves in a circular path of radius $1.6\times 1{0}^{−5}\text{ }\text{m}$. The magnetic field ($B$) is:

(Shift II Memory Based)

YESPYQ · Accepted Answer

Correct answer: b) $1.3\times 1{0}^{5}\text{ }\text{T}$. When the electric field is switched off, the magnetic force provides the centripetal force required for circular motion. The magnetic force acting on the proton is given by: ${\mathrm{F}}_{\mathrm{m}}=\mathrm{qvB}\sin 90^\circ =\mathrm{qvB}$ ..... (1) The centripetal force is given by $\mathrm{Fc}=\frac{{\mathrm{mv}}^{2}}{\mathrm{r}}$ .... (2) By equating equation (1) and (2), we get $\mathrm{qvB}=\frac{{\mathrm{mv}}^{2}}{\mathrm{r}}\ \mathrm{B}=\frac{\mathrm{m}\mathrm{v}}{\mathrm{r}\mathrm{q}}\ \mathrm{B}=\frac{1.67\times {10}^{-27}\times 2\times {10}^{8}}{1.6\times {10}^{-5}\times 1.6\times {10}^{-19}}\ \mathrm{B}=1.3\times {10}^{5}\mathrm{T}$

A proton is moving with a uniform velocity of \(2\times 1{0}^{8}\text{ }\text{m/s}\) in uniform magnetic and electric fi…

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