A straight long wire carrying a current of 30 A is placed in an external uniform magnetic field of \( 4\times10^{-4}\,\mathrm{T} \) parallel to the current. The magnitude of the resultant magnetic field at a point 2 cm away from the wire is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5\times10^{-4}\,\mathrm{T} \)
The field of a long straight current forms circles around the wire, so at any point it is perpendicular to the wire. The external field is parallel to the wire, so the two fields are at right angles and are added as perpendicular vectors.
\( \displaystyle B_{\mathrm{wire}} = \frac{\mu_0 I}{2\pi r} = \frac{2\times10^{-7} \times 30}{2\times10^{-2}} = 3\times10^{-4}\,\mathrm{T} \)
\( \displaystyle B = \sqrt{B_{\mathrm{wire}}^{2} + B_{\mathrm{ext}}^{2}} = \sqrt{(3\times10^{-4})^{2} + (4\times10^{-4})^{2}} = 5\times10^{-4}\,\mathrm{T} \)
Shortcut: The numbers 3 and 4 form the standard right triangle 3-4-5, so the resultant is \( 5\times10^{-4}\,\mathrm{T} \) without any calculation.