A wire carrying a current I along the positive x -axis has length L . It is kept in a magnetic field B = (2i + 3j - 4k) T . The magnitude of the magne - Physics MCQ NEET 2023 2026 | ExamTest.live

hardMCQNEET 20232026Physics

1 mark

A wire carrying a current I along the positive x -axis has length L . It is kept in a magnetic field B = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) T . The magnitude of the magnetic force acting on the wire is:

Solution & Step-by-step Explanation

The force on a current-carrying wire is F = I (L \times B) .Length vector L = L \hat{i} . L \times B = (L \hat{i}) \times (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) L \times B = L (2 \hat{i} \times \hat{i} + 3 \hat{i} \times \hat{j} - 4 \hat{i} \times \hat{k}) L \times B = L (0 + 3 \hat{k} + 4 \hat{j}) = L (4 \hat{j} + 3 \hat{k}) Magnitude: ∣ F ∣ = I \times L 4^{2} + 3^{2} = I \times L 16 + 9 = 5 I L

Try it yourself before checking the explanation above.

A wire carrying a current I along the positive x -axis has length L . It is kept in a magnetic field B = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) T . The magnitude of the magnetic force acting on the wire is:

3 I L

5 I L

5 I L

3 I L

Practice with timed mock tests and track your performance across Physics.