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# The magnetic field density applied in a cyclotron is $3.5 \;T$ what will be the frequency of electric field that must be applied between the dees in order to accelerate protons. Mass of protons $= 1.67 \times 10^{27}\;kg$

$\begin{array}{1 1} 5.35 \times 10^7 \;Hz \\ 535 \times 10^7 \;Hz \\ 535 \times 10^7 \;Hz \\ 0.535 \times 10^7 \;Hz \end{array}$

Given $B= 3.5 \;T$
$q= 1.6 \times 10^{-19}$
$m= 1.67 \times 10^{-27}$
$v= \large\frac{B_q}{2 \pi m} =\frac{3.5 \times 1.6 \times 10^{-19}}{2 \times \Large\frac{22}{7} \times 1.67 \times 10^{27}}$
$\qquad= 5.35 \times 10^7 \;H^z$
Answer : $5.35 \times 10^7 \;Hz$