Question:
A long straight wire carries a current of 10 A. An electron travels with velocity \(5 \times 10^6\,m\,s^{-1}\) at a distance 0.1 m from it, in a direction opposite to the current. Estimate the force experienced by the electron.
A long straight wire carries a current of 10 A. An electron travels with velocity \(5 \times 10^6\,m\,s^{-1}\) at a distance 0.1 m from it, in a direction opposite to the current. Estimate the force experienced by the electron.
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For magnetic force problems, always check orientation to decide whether \(\sin\theta = 1\) or not.
Updated On: Jun 19, 2026
- \(1.6 \times 10^{-18}\,N\)
- \(1.6 \times 10^{-17}\,N\)
- \(1.6 \times 10^{-16}\,N\)
- \(1.6 \times 10^{-15}\,N\)
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The Correct Option is B
Solution and Explanation
Step 1: Magnetic field due to long straight wire.
\[ B = \frac{\mu_0 I}{2\pi r} \]
Step 2: Substitute values.
\[ B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.1} \]
Step 3: Simplify.
\[ B = \frac{4 \times 10^{-6}}{0.2} = 2 \times 10^{-5}\,T \]
Step 4: Force on moving charge.
\[ F = qvB \sin\theta \] Here electron moves opposite to current ⇒ velocity ⟂ magnetic field ⇒ \(\sin\theta = 1\).
Step 5: Substitute values.
\[ F = (1.6 \times 10^{-19})(5 \times 10^6)(2 \times 10^{-5}) \]
Step 6: Final calculation.
\[ F = 1.6 \times 10^{-17}\,N \]
\[ B = \frac{\mu_0 I}{2\pi r} \]
Step 2: Substitute values.
\[ B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.1} \]
Step 3: Simplify.
\[ B = \frac{4 \times 10^{-6}}{0.2} = 2 \times 10^{-5}\,T \]
Step 4: Force on moving charge.
\[ F = qvB \sin\theta \] Here electron moves opposite to current ⇒ velocity ⟂ magnetic field ⇒ \(\sin\theta = 1\).
Step 5: Substitute values.
\[ F = (1.6 \times 10^{-19})(5 \times 10^6)(2 \times 10^{-5}) \]
Step 6: Final calculation.
\[ F = 1.6 \times 10^{-17}\,N \]
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