A rectangular loop measures 10 cm by 5 cm. It moves along its longer side into a region of uniform magnetic field of 0.6 T. The field is normal to the plane of the loop, and there is no field outside the region. When 4 cm of its longer side is inside the field, the magnetic flux through the loop is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.2\times10^{-3}\ \mathrm{Wb}\)
Flux is counted only over the part of the loop that lies in the field. The part still outside has no field through it, so it adds nothing. The field is along the normal to the loop, so \( \theta=0^\circ \).
The part inside the field is 4 cm long and 5 cm wide.
\( \begin{aligned} A &= 4\times5\ \mathrm{cm^{2}} \\ &= 20\ \mathrm{cm^{2}} \\ &= 2\times10^{-3}\ \mathrm{m^{2}} \end{aligned} \)
\( \begin{aligned} \Phi_B &= BA\cos\theta \\ &= 0.6\times2\times10^{-3}\times\cos0^\circ \\ &= 1.2\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
NCERT (2026-27):
Magnetic Flux – “… Magnetic flux through a plane of area A placed in a uniform magnetic field B … can be written as”
\( \Phi_B=\mathbf{B}\cdot\mathbf{A}=BA\cos\theta \)\( \qquad (6.1) \)
“where θ is angle between B and A. …”