A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is B = (3i+4k)T. The quantity of flux through the loop ABCDEFA (in Wb) is _______. Numerical answer type. Work it out, then open the answer.

Answer: 175. Split the loop into two squares: ABCD in the z=0 plane and ADEF in the x=0 plane, each of area 25 m ². Going A→B→C→D, the area vector of ABCD is 25k; going D→E→F→A, the area vector of ADEF is 25i. A = (25i + 25k) m ². φ = B = 3(25) + 4(25) = 175 Wb.

A loop ABCDEFA of straight edges has six corner points A(0, 0, 0)… | JEE Main 2020 (7 Jan, Shift 1)

Q72JEE Main 2020 · 7 JAN · SHIFT 1Difficult

Magnetic Flux & Faraday's Law

A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is $\vec{B} = \left(3\hat{i}+4\hat{k}\right)T$ . The quantity of flux through the loop ABCDEFA (in Wb) is _______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 175

Split the loop into two squares: ABCD in the $z=0$ plane and ADEF in the $x=0$ plane, each of area $25$ m $^2$ .
Going A→B→C→D, the area vector of ABCD is $25\hat{k}$ ; going D→E→F→A, the area vector of ADEF is $25\hat{i}$ .
$\vec{A} = (25\hat{i} + 25\hat{k})$ m $^2$ .
$\phi = \vec{B}\cdot\vec{A} = 3(25) + 4(25) = 175$ Wb.

Report a mistake

Magnetic Flux & Faraday's Law question from JEE Main 2020 (7 Jan, Shift 1)

More questions on Magnetic Flux & Faraday's Law