A) Mass
B) Speed
C) Charge
D) Magnetic field
Correct Answer: B
Solution :
\[\omega =\frac{2\pi }{T}=\frac{qB}{m}\]Þ \[\omega \propto v{}^\circ \] \[\left( \because T=\frac{2\pi m}{qB} \right)\]You need to login to perform this action.
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