The oscillator frequency of a cyclotron is $10 \,MHz$. What should be the operating magnetic field for accelerating proton is ......... $T$
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(d)

$f=\frac{q B}{2 \pi m} \quad\left(\because T =\frac{2 \pi m}{q B}\right)$

$\Rightarrow 10 \times 10^6=\frac{1.6 \times 10^{-19} \times B}{2 \pi\left(1.6 \times 10^{-27}\right)}$

$B=0.656 \,T$

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