A charge \( q = 1.6 \times 10^{-12} \text{ C} \) moving with speed of \( v \text{ m s}^{-1} \) crosses electric field \( |\vec{E}| = 6 \times 10^4 \text{ V m}^{-1} \) and magnetic field \( |\vec{B}| = 1.2 \text{ T} \). The electric field and magnetic fields are crossed and velocity \( v \) is also perpendicular to both. If the charge particle crosses both fields undeflected, the value of \( v \) is
Solution:
For a particle to cross perpendicular electric and magnetic fields undeflected, the net force must be zero, which requires \( qE = qvB ⇒ v = \frac{E}{B} \). Substituting the given values: \( v = \frac{6 \times 10^4}{1.2} = 5 \times 10^4 \text{ m/s} \).
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