(B) For the electron to move undeflected,the net Lorentz force must be zero.$F_{net} = F_e + F_m = 0$$qE = qvB$$E = vB$Since kinetic energy $KE = \fra

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(B) For the electron to move undeflected,the net Lorentz force must be zero.$F_{net} = F_e + F_m = 0$$qE = qvB$$E = vB$Since kinetic energy $KE = \frac{1}{2}mv^2$,the velocity $v = \sqrt{\frac{2 \times KE}{m}}$.Substitute $KE = 5 \ eV = 5 \times 1.6 \times 10^{-19} \ J$ and $m = 9 \times 10^{-31} \ kg$:$v = \sqrt{\frac{2 \times 5 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16}{9} \times 10^{12}} = \frac{4}{3} \times 10^6 \ m/s$.Now,calculate $E = vB$ with $B = 3 \ \mu T = 3 \times 10^{-6} \ T$:$E = (\frac{4}{3} \times 10^6) \times (3 \times 10^{-6}) = 4 \ N C^{-1}$.

Class 12 Physics Moving Charges and Magnetism Motion of Charged Particle In Magnetic Field

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An electron with kinetic energy $5 \ eV$ enters a region of uniform magnetic field of $3 \ \mu T$ perpendicular to its direction. An electric field $E$ is applied perpendicular to the direction of velocity and magnetic field. The value of $E$,so that the electron moves along the same path,is . . . . . $N C^{-1}$.
(Given: mass of electron $= 9 \times 10^{-31} \ kg$,electric charge $= 1.6 \times 10^{-19} \ C$)

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Moving Charges and Magnetism

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Motion of Charged Particle In Magnetic Field

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An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.

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$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then:

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