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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2024
  • Hard
  • Numerical

A charge of \(4.0 \mu \mathrm{C}\) is moving with a velocity of \(4.0 \times 10^6 \mathrm{~ms}^{-1}\) along the positive \(\mathrm{y}\)-axis under a magnetic field \(\vec{B}\) of strength \((2 \hat{k})\  T\). The force acting on the charge is \(x \hat{i} N\). The value of \(x\) is _______.

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Answer & Solution

Correct Answer

(D) \(32\)

Step-by-step Solution

Detailed explanation

\(\mathrm{q}  =4 \mu \mathrm{C}, \overrightarrow{\mathrm{v}}=4 \times 10^6\)     

\( \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} \)

\(\overrightarrow{\mathrm{B}}  =2 \hat{\mathrm{k}} \mathrm{T} \)
\(\overrightarrow{\mathrm{F}}  =\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}) \)
\( =4 \times 10^{-6}\left(4 \times 10^6 \hat{\mathrm{j}} \times 2 \hat{\mathrm{k}}\right) \)
\( =4 \times 10^{-6} \times 8 \times 10^6 \hat{\mathrm{i}} \)
\(\overrightarrow{\mathrm{F}}  =32 \hat{\mathrm{i}} \mathrm{N} \)
\(\mathrm{x}  =32\)

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