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Given\begin{array}{l}k \cdot f=1.20 \mathrm{kCV} \Rightarrow 1.2 \times 10^{3} \times 1.6 \times 10^{-19} \mathrm{~J} \\R=25 \mathrm{~cm} \Rightarrow 25 \times 10^{-2} \mathrm{~m}\end{array}(A) we know that\begin{aligned}& 1.2 \times 1.6 \times 10^{-16}=\frac{1}{2} \times 9.1 \times 10^{-31} \times \mathrm{v}^{2} \\V & =\frac{1}{2} m_{e} v^{0.42 \times 10^{15}} \\V & {\left[m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right] } \\\omega e \quad \text { know } & =2.04 \times 10^{7} \mathrm{~m} / \mathrm{s} \\W & =\frac{2.04 \times 10^{7}}{25 \times 10^{-2}} \\\omega & =0.08 \times 10^{9} \\\omega & =8 \times 10^{7} \mathrm{~m} / \mathrm{s}\end{aligned}(b) magnitude of Magnetic field:we also know\begin{array}{l}B=0.46 \times 10^{-3} \mathrm{~T} \times 10^{-2} \quad R=25 \times 10^{-2} \mathrm{~m} \\\Rightarrow B=4.6 \times 10^{-4} \mathrm{~T} \\\end{array}If we Incrcase Kinetic Energy (2factrr)\omega \cdot \epsilon .=2 k . \epsilon \text {. }K. E .=\frac{1}{2} m_{e}(\omega R)^{2} in creale by factor' 2 )if you have any doubt about answer please ask in comment section Thank you dear ...