A charged particle moving along a straight line path enters a uniform magnetic field of 4 mT at right angles…

A charged particle moving along a straight line path enters a uniform magnetic field of 4 mT at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7 \mathrm{C} \mathrm{kg}^{-1}$, the angular velocity of the particle in the magnetic field is
  1. $64 \times 10^4 \mathrm{rad} \mathrm{s}^{-1}$
  2. $32 \times 10^4 \mathrm{rad} \mathrm{s}^{-1}$
  3. $16 \times 10^4 \mathrm{rad} \mathrm{s}^{-1}$
  4. $48 \times 10^4 \mathrm{rad} \mathrm{s}^{-1}$

Solution

$B=4 \mathrm{mT}=4 \times 10^{-3} \mathrm{~T}, \frac{\mathrm{q}}{\mathrm{m}}=8 \times 10^7 \mathrm{c} \mathrm{kg}^{-1}$
$\therefore \quad$ Angular velocity, $\mathrm{w}=\frac{\mathrm{B}_{\mathrm{q}}}{\mathrm{m}}=4 \times 10^{-3} \times 8 \times 10^7$ $=32 \times 10^4 \mathrm{rad} \mathrm{s}^{-1}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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