One wire, lying on the x-axis, carries a current of 8.0 A in the positive x-direction. Another wire, lying on the y-axis, carries a current of 12 A in the positive y-direction. What is the magnitude of the magnetic field at (x, y) = (8.0 cm, 12.0 cm)? A) 1.3 × 10⁻⁵ T B) 3.0 × 10⁻⁵ T C) 1.7 × 10⁻⁵ T D) 4.3 × 10⁻⁵ T E) 4.0 × 10⁻¹⁰ T

Question

Quizplus · Accepted Answer

We'll use the formula for magnetic field due to a current-carrying wire:
B = μ₀I/(2πr)
where μ₀ is the magnetic constant (4π×10⁻⁷ T*m/A), I is the current, and r is the distance from the wire.

For the x-axis wire at (8.0 cm, 0), the distance to (8.0 cm, 12.0 cm) is r = 12.0 cm. So the magnetic field due to this wire is:
B₁ = μ₀I₁/(2πr) = (4π×10⁻⁷ T*m/A)×(8.0 A)/(2π×0.12 m) = 4.2×10⁻⁴ T

For the y-axis wire at (0, 12.0 cm), the distance to (8.0 cm, 12.0 cm) is r = 8.0 cm. So the magnetic field due to this wire is:
B₂ = μ₀I₂/(2πr) = (4π×10⁻⁷ T*m/A)×(12 A)/(2π×0.08 m) = 2.4×10⁻⁴ T

Since the two wires are perpendicular to each other, the magnetic fields due to them are also perpendicular. Therefore, we can find the total magnetic field at (8.0 cm, 12.0 cm) using the Pythagorean theorem:
B = √(B₁² + B₂²) = √[(4.2×10⁻⁴ T)² + (2.4×10⁻⁴ T)²] = 1.7×10⁻⁵ T

So the answer is C, 1.7 × 10⁻⁵ T.

One Wire, Lying on the X-Axis, Carries a Current of 8.0