Q6. A square shaped wire loop of side L is carrying a current I. What is the magnetic field at the p

Q: What is the correct answer for CUET UG Physics Q6?

The correct answer is b. The magnetic field at the center of a square loop carrying current I is the vector sum of the magnetic fields produced by each of its four sides. For a single straight current-carrying segment, the magnetic field at a perpendicular distance 'x' from its midpoint is given by B_segment = (μ₀I / (4πx)) * (sinθ₁ + sinθ₂). For a square of side L, the distance 'x' from the center to the midpoint of any

Q: Q6. A square shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop? (α) μοι πι√2 2√2 μοι (b) πι (c)…

The correct answer is b. The magnetic field at the center of a square loop carrying current I is the vector sum of the magnetic fields produced by each of its four sides. For a single straight current-carrying segment, the magnetic field at a perpendicular distance 'x' from its midpoint is given by B_segment = (μ₀I / (4πx)) * (sinθ₁ + sinθ₂). For a square of side L, the distance 'x' from the center to the midpoint of any

Q: What are the options for this CUET UG Physics question?

A. μοι / (πι√2) · B. 2√2 μοι / (πι) · C. 2μοι / (πι) · D. √2 μοι / (πι)

Question

Q6. A square shaped wire loop of side L is carrying a current I. What is the magnetic field at the point of intersection of diagonals of the square wire loop?
(α) μοι
πι√2
2√2 μοι
(b) πι
(c) 201
πι
√2 μοι
(d) πι

FirstInTest · Accepted Answer

The magnetic field at the center of a square loop carrying current I is the vector sum of the magnetic fields produced by each of its four sides. For a single straight current-carrying segment, the magnetic field at a perpendicular distance 'x' from its midpoint is given by B_segment = (μ₀I / (4πx)) * (sinθ₁ + sinθ₂). For a square of side L, the distance 'x' from the center to the midpoint of any side is L/2. The angles θ₁ and θ₂ subtended by the ends of the wire segment at the center, with respect to the perpendicular line, are both 45° (since tanθ = (L/2)/(L/2) = 1). So, sin45° = 1/√2. Therefore, for one side, B_side = (μ₀I / (4π(L/2))) * (1/√2 + 1/√2) = (μ₀I / (2πL)) * (2/√2) = (μ₀I√2) / (2πL). Since there are four identical sides, and their fields all point in the same direction at the center (by right-hand rule), the total magnetic field B_total = 4 * B_side = 4 * (μ₀I√2) / (2πL) = (2√2 μ₀I) / (πL). Options (a), (c), and (d) result from incorrect calculations or formulas for the magnetic field of a current loop.

Answer

μοι / (πι√2)

Answer

2√2 μοι / (πι)

Answer

2μοι / (πι)

Answer

√2 μοι / (πι)

Related Questionsin Physics

More from CUET UG

Browse CUET Questions by Subject

Practice More CUET UG Questions