JEE Main 2024 · 6 April · morning · Single correct · +4 / −1
An element Δ l=Δ xî is placed at the origin and carries a large current I=10 A. The magnetic field on the y-axis at a distance of 0.5 m from the elements Δ x of 1 cm length is:
- (a) 10 × 10⁻⁸ T
- (b) 8 × 10⁻⁸ T
- (c) 4 × 10⁻⁸ T
- (d) 12 × 10⁻⁸ T
Answer and solution
Answer (c) 4 × 10⁻⁸ T
Solution
Physics sir · 4 steps
A tiny current element means Biot-Savart directly, and here the angle is a clean 90 degrees.
- Use Biot-Savart. The element is only 1 cm long, much shorter than the 0.5 m distance to P. So treat it as a single current element and apply the Biot-Savart law once.
dB = μ₀/4π (I Δ x sin θ)/r²
- Find the angle. The element points along î and P lies along ĵ from it, so θ = 90° and sin θ = 1.
- Put in numbers. With μ₀/4π = 10⁻⁷ T m A⁻¹, I = 10 A, Δ x = 10⁻² m and r² = 0.25 m², you get dB = (10⁻⁷ × 10 × 10⁻²)/0.25 = 4 × 10⁻⁸ T.
- So. The field at P is 4 × 10⁻⁸ T, directed along î × ĵ = k̂. That is option (c).
The idea to keep. For a short current element, use μ₀/4π(I dl sin θ)/r². Find the angle between the element and the line to the point, and convert cm to m first.
Answer from the NTA final answer key. Solution by PrepWiser.