Q118 / 120 · VITEEE 2019 · Physics

The work done by an uniform magnetic field, on a moving charge is

  • A
    zero because $\vec{F}$ acts parallel to $\overrightarrow{\mathrm{v}}$
  • B
    positive because $\vec{F}$ acts perpendicular to $\overrightarrow{\mathrm{v}}$
  • Correct answer
    C
    zero because $\vec{F}$ acts perpendicular to $\overrightarrow{\mathrm{v}}$
  • D
    negative because $\vec{F}$ acts parallel to $\overrightarrow{\mathrm{v}}$

Solution

Force on moving charge while moving in magnetic field is; $\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ where $\overrightarrow{\mathrm{F}}$ is perpendicular to $\overrightarrow{\mathrm{V}}$. Work done/sec $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{v}}=\mathrm{Fv} \cos 90^{\circ}=0$

Hence, the correct answer is option: 'C'.

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