An electron of kinetic energy of 7.2×10^-18j is revolving in circular path in magnetic field 9×10^-5wb/m2 then radius of its circular path is
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redius = underroot 2mEk/ QB
= underroot 2*9.1*10^-31*7.2*10^-18 /1.6*10^-19*9*10^-5
r = approx 10^-12
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~* THE ANSWER OF YOUR QUESTION IS..*~
✍✍✍✍✍✍✍✍________________________________________
redius = underroot 2mEk/ QB
= underroot 2*9.1*10^-31*7.2*10^-18 /1.6*10^-19*9*10^-5
r = approx 10^-12
HOPE IT WILL HELP YOU
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