Ncert Exercises & Solutions

A loop, made of straight edges has six corners at A(0, 0, 0), B(L, 0, 0), C(L, L, 0), D(0, L, 0), E(0, L, L) and F(0, 0, L). A magnetic field B= $B_{0} (\hat{i} + \hat{k})$ T is present in the region. The flux passing through the loop ABCDEFA (in that order) is:

1. B₀L² Wb³

2. 2B₀L² Wb

3. $\sqrt2$ B₀L² Wb

4. 4B₀L² Wb

(b) Hint: The flux passing through the loop depends on the area of the loop.

Step 1: Find the flux.

The magnetic flux linked with the surface of area A in a uniform magnetic field is given by;

$ϕ = \vec{B} . \vec{A}$

$\vec{A} = {\vec{A}}_{1} + {\vec{A}}_{2} = (L^{2} \hat{k} + L^{2} \hat{i})$

$and \vec{B} = B_{0} (\hat{i} + \hat{k}) T$

$Now ϕ = \vec{B} . \vec{A} = B_{0} (\hat{i} + \hat{k}) . (L^{2} \hat{k} + L^{2} \hat{i})$

$= 2 B_{0} L^{2} Wb$

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