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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

एक आवेश \(q\) पर \(v\) वेग से गतिमान होने पर लगने वाला स्थिरविद्युत बल \((\vec{F}_1)\) और चुंबकीय बल \((\vec{F}_2)\) को _______ लिखा जा सकता है।

  1. A \(\vec{F}_1=q \vec{V} \cdot \vec{E}, \vec{F}_2=q(\vec{B} \cdot \vec{V})\)
  2. B \(\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{V}})\)
  3. C \(\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{E}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\)
  4. D \(\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{V}})\)
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Answer & Solution

Correct Answer

(C) \(\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{E}}, \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\)

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Detailed explanation

\(\overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{E}}\) (सिद्धांत) \(\overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})\)
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