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(Solved): An electron with a kinetic energy of \( 1.09 \times 10^{4} \mathrm{eV} \) is initially moving sout ...
An electron with a kinetic energy of \( 1.09 \times 10^{4} \mathrm{eV} \) is initially moving south in a uniform magnetic field of magnitude \( 548 \mathrm{G} \), directed east. Describe in detail the electron's trajectory, including its shape and orientation. the radius of the path direction of initial force shape of trajectory
Expert Answer
Solution : given , KE of electron = 1.09*10^4 eV and is moving south. magnetic field (B) = 548 Gauss = 548 * 10^-4 T = 0.0548 T mass of electron (m) =