A hemispherical bowl of radius 10 cm is closed by a flat circular lid. This closed surface is kept in a uniform magnetic field of 0.2 T. The field is normal to the lid. The magnitude of the magnetic flux through the curved surface of the bowl is (take \(\pi=3.14\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(6.28\times10^{-3}\ \mathrm{Wb}\)
The bowl and its lid together form a closed surface. By Gauss’s law for magnetism, the net magnetic flux through it is zero. So the flux leaving through the curved part is equal in size to the flux entering through the lid.
\( r=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( B=0.2\ \mathrm{T} \)
The lid is a flat circle normal to the field, so its flux is \( B\pi r^{2} \).
\( \begin{aligned} |\phi_{\text{curved}}| &= |\phi_{\text{lid}}| \\ &= B\pi r^{2} \\ &= 0.2\times3.14\times(0.1)^{2} \\ &= 6.28\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
NCERT (2026-27):
Magnetism and Gauss’s Law – “Thus, Gauss’s law for magnetism is: The net magnetic flux through any closed surface is zero.”