One proton beam enters a magnetic field of ${10^{ - 4}}$ $T$ normally, Specific charge = ${10^{11}}\,C/kg.$ velocity = ${10^7}\,m/s$. What is the radius of the circle described by it....$m$
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(b) $r = \frac{{mv}}{{qB}} = \frac{{{{10}^7}}}{{{{10}^{11}} \times {{10}^{ - 4}}}} = 1m\;(\;q/m = {10^{11}}C/kg)$
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