Question

A proton is travelling along the X-direction with velocity \(5 \times 10^6\,ms^{-1}\). The magnitude of force experienced by the proton in a magnetic field \(\vec{B} = (0.2\hat{i} + 0.4\hat{k})\) tesla is

• A. \(3.2 \times 10^{-13} N\)
• B. \(5.3 \times 10^{-12} N\)
• C. \(3.2 \times 10^{13} N\)
• D. \(6.3 \times 10^{-13} N\)
• E. \(3.5 \times 10^{-12} N\)

Hint:

Only perpendicular components contribute in cross product.

Answer 1

The Correct Option is A

Concept: Magnetic force on moving charge: \[ \vec{F} = q (\vec{v} \times \vec{B}) \]

Step 1: Velocity vector. \[ \vec{v} = 5 \times 10^6 \hat{i} \]

Step 2: Magnetic field vector. \[ \vec{B} = 0.2\hat{i} + 0.4\hat{k} \]

Step 3: Evaluate cross product. \[ \vec{v} \times \vec{B} = (5\times10^6 \hat{i}) \times (0.2\hat{i} + 0.4\hat{k}) \] Using rules: \[ \hat{i} \times \hat{i} = 0, \quad \hat{i} \times \hat{k} = -\hat{j} \] \[ \vec{v} \times \vec{B} = 5\times10^6 \times 0.4 (-\hat{j}) \] \[ = -2\times10^6 \hat{j} \]

Step 4: Multiply by charge. \[ F = qvB = (1.6\times10^{-19})(2\times10^6) \]

Step 5: Final calculation. \[ F = 3.2\times10^{-13} N \]

Step 6: Conclusion. \[ \boxed{3.2\times10^{-13} N} \]