A square loop of side 20 cm lies in the plane of the page. A uniform magnetic field of 0.3 T acts over its left half and points into the page. A uniform field of 0.1 T acts over its right half and points out of the page. The magnitude of the total magnetic flux through the loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4\times10^{-3}\ \mathrm{Wb}\)
When the field is not the same over a surface, the flux through each part is found and the results are added. Take the normal to the loop pointing into the page. The left field is along this normal, so \( \theta=0^\circ \). The right field is opposite to it, so \( \theta=180^\circ \) and its flux is negative.
Area of each half:
\( \begin{aligned} A &= \frac{0.20\times0.20}{2} \\ &= 0.02\ \mathrm{m^{2}} \end{aligned} \)
Total flux through the loop:
\( \begin{aligned} \Phi_B &= B_1A\cos0^\circ+B_2A\cos180^\circ \\ &= 0.3\times0.02-0.1\times0.02 \\ &= 4\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
NCERT (2026-27):
Magnetic Flux – “If the magnetic field has different magnitudes and directions at various parts of a surface … then the magnetic flux through the surface is given by”
\( \displaystyle \Phi_B=\mathbf{B}_1\cdot d\mathbf{A}_1+\mathbf{B}_2\cdot d\mathbf{A}_2+\dots=\sum_{all}\mathbf{B}_i\cdot d\mathbf{A}_i \)\( \qquad (6.2) \)