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A beam of proton with a velocity of 4xx1...

A beam of proton with a velocity of `4xx10^(5)ms^(-1)` enters a uniform magnetic field of 0.3 T at an angle of `60^(@)` to the magnetic field/The radius of helical path taken by proton beam is

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Radius of helix =`mV sin theta/qb`= `(1.67xx10^(-27)) (4xx10^(5)) (sin60^(@))/(1.6 xx 10^(-19)) (0.3)=12.05 xx 10^(-3)`m
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VMC MODULES ENGLISH-MOVING CHARGES & MAGNETISM -Illustration
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