Question

what is the radios of electron moving at a speed of 3*10power7 m/s in a magnetic field of 6*10power-4 T perpendicular to it. what is its frequency? ​

Answer 1

Answer:

Solution :

Here, m=9×10−31kg,  

q=1⋅6×10−19C, v=3×107ms−1,  

B=6×10−4T  

r=mvqB=(9×10−31)×(3×107)(1⋅6×10−19)(6×10−4)=0⋅28m  

v=v2πr=Bq2πm=(6×10−4)×(1⋅6×10−19)2×(22/7)×9×10−31  

=1⋅7×107Hz  

Ek=12mv2=12×(9×10−31)×(3×107)2J  

=40⋅5×10−17J=40⋅5×10−171⋅6×10−16keV  

=2⋅53keV

Explanation:

Hope helps

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