Question Please solve this question #3. [5 points] An electron of kinetic energy 1.20 keV circles in a plane perpendicular to a uniform magnetic field. The radius of the orbit is 25.0 cm. (a) Find the magnitude of the magnetic field and the angular speed of the electron (b) What would happen to the angular speed if the energy of the electron is increased by a factor of 2?

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 ...