A rectangular loop of sides 20 cm and 10 cm is placed in a uniform magnetic field of 0.5 T. The plane of the loop makes an angle of \(30^\circ\) with the direction of the field. The magnetic flux through the loop is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(5.0\times10^{-3}\ \mathrm{Wb}\)
The flux through a flat loop in a uniform field is \( \Phi_B=BA\cos\theta \). The angle \( \theta \) is measured between the field and the area vector, which is normal to the plane. The plane is at \( 30^\circ \) to the field, so the normal is at \( 90^\circ-30^\circ=60^\circ \) to it.
\( B=0.5\ \mathrm{T} \)
\( A=0.02\ \mathrm{m^{2}} \) (20 cm by 10 cm)
\( \theta=60^\circ \)
\( \begin{aligned} \Phi_B &= BA\cos\theta \\ &= 0.5\times0.02\times\cos60^\circ \\ &= 5.0\times10^{-3}\ \mathrm{Wb} \end{aligned} \)
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.”