Question

A long wire lies along \(x\)-axis and carries a current of \(40\,\text{A}\) in positive \(x\)-direction. A second long wire is perpendicular to the \(xy\)-plane, passes through point \((3.0\,\text{m})\hat{j}\) and carries a current along positive \(z\)-direction. If the magnitude of resultant magnetic field at the point \((2.0\,\text{m})\hat{j}\) is \(5\times 10^{-6}\,\text{T}\), then the current in the second wire is \((\mu_0=4\pi\times 10^{-7}\,\text{SI unit})\)

• A. \(30\,\text{A}\)
• B. \(15\,\text{A}\)
• C. \(25\,\text{A}\)
• D. \(7.5\,\text{A}\)

Hint:

For a long straight current-carrying wire, \[ B=\frac{\mu_0 I}{2\pi r}. \] If two magnetic fields are perpendicular, use \[ B_{\text{net}}=\sqrt{B_1^2+B_2^2}. \]

Answer 1

The Correct Option is B

Step 1: Compute field from first wire.
B₁ = μ₀I₁/(2πr₁) = (4π×10⁻⁷×40)/(2π×2) = 4×10⁻⁶ T.

Step 2: Express field from second wire in terms of unknown I.
B₂ = μ₀I/(2πr₂) = (4π×10⁻⁷×I)/(2π×1) = 2×10⁻⁷ I.

Step 3: Use Pythagorean addition (fields perpendicular).
B_net² = B₁² + B₂² → (5×10⁻⁶)² = (4×10⁻⁶)² + B₂² → B₂ = 3×10⁻⁶ T.

Step 4: Solve for I.
3×10⁻⁶ = 2×10⁻⁷ I → I = 15 A.

Step 5: Final Answer:
15 A.