JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Electromagnetic Induction

A loop

A B C D E F A

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 in

B = (3 \hat{i} + 4 \hat{k}) T

. The quantity of flux through the loop ABCDEFA (in Wb ) is ____.

Answer: 175

Solution

The closed loop is made of two flat squares of side

5 m

: one lying in the

x y

-plane with area vector

25 \hat{k}

, and one lying in the

y z

-plane with area vector

25 \hat{i}

, giving a net surface

A = 25 \hat{i} + 25 \hat{k} m^{2}

. The flux is

ϕ = B \cdot A = (3 \hat{i} + 4 \hat{k}) \cdot (25 \hat{i} + 25 \hat{k}) = 75 + 100 = 175 Wb

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Integer answer

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