Velocity Selector And Crossed Fields

Question

In a region with a uniform electric field of $3 	imes 10^{4}$ V/m perpendicular to a magnetic field of $0.06$ T, a charged particle passes undeflected; what is the speed of this particle?

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NCERT Class 12, Chapter 4 (Moving Charges and Magnetism) describes a velocity selector in which mutually perpendicular electric and magnetic fields are arranged so that their forces on a moving charge oppose each other. A particle passes undeflected only when the electric force balances the magnetic force: $qE = qvB$, which is independent of charge and gives the selected speed $v = \dfrac{E}{B}$. Substituting $E = 3 	imes 10^{4}$ V/m and $B = 0.06$ T: $v = \dfrac{3 	imes 10^{4}}{0.06} = 5 	imes 10^{5}$ m/s. The value $1.8 	imes 10^{3}$ m/s comes from wrongly multiplying $E$ by $B$. The value $2 	imes 10^{6}$ m/s and $5 	imes 10^{6}$ m/s result from decimal-place errors in dividing by $0.06$. Plausibility check: units give $\dfrac{	ext{V/m}}{	ext{T}} = \dfrac{	ext{N/C}}{	ext{N·s/(C·m)}} = 	ext{m/s}$, correctly a speed, and $5 	imes 10^{5}$ m/s is a realistic selected speed for charged particles in such an apparatus. Note that because both the electric and magnetic forces scale with the same charge $q$, the balance condition is independent of the sign and magnitude of the charge, so the selector transmits the chosen speed regardless of the particle species; this charge-independence is the very feature that makes it a clean velocity filter feeding a mass spectrometer.

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Velocity Selector And Crossed Fields

Select the correct option:

🔒 Solution Hidden from View

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About This Question

Solution

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