Question

Please solve the above problem..​

Answer 1

Explanation:

$\tt Frequency\:(f) = 12\:MHz = 12 × {10}^{6} \: Hz \\ \tt Dee\:Radius\:(R) = 70\:cm = 0.7\:m \\ \tt Mass\:of\:electron\:(m) =9.1 × {10}^{-31} \:kg \\ \\ \tt We\:know\:that \\ \\ \tt f = \frac{qB}{2\pi m} \\ \\ \tt \longrightarrow B = \frac{2\pi mf}{q} \\ \\ \tt \longrightarrow B = \frac{2×3.14×9.1×{10}^{-31} × 12 ×{10}^{6}}{1.6×{10}^{-19}} \\ \\ \tt \longrightarrow B = 4.28 × {10}^{-4} \\ \\ \tt \boxed{\bold{\underline{\red {\tt Operating\:magnetic\:field\:for\: accelerating\:electron\:is\:4.28×{10}^{-4}\:Tesla}}}}$

Answer 2

Answer:

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