A square coil of 50 turns and side 10 cm is placed in a uniform magnetic field of 0.4 T. At first, the normal to the coil is along the field. The coil is then turned so that its normal makes an angle of \(60^\circ\) with the field. The magnitude of the change in the total flux linked with the coil is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(0.1\ \mathrm{Wb}\)
The flux through one turn is \( \Phi_B=BA\cos\theta \), where \( \theta \) is the angle between the field and the normal to the coil. For \( N \) turns, the flux linked with the coil is \( N\Phi_B \). The side is 0.1 m, so \( A=0.01\ \mathrm{m^{2}} \).
At first, \( \theta=0^\circ \):
\( \begin{aligned} N\Phi_1 &= NBA\cos0^\circ \\ &= 50\times0.4\times0.01\times1 \\ &= 0.2\ \mathrm{Wb} \end{aligned} \)
After turning, \( \theta=60^\circ \):
\( \begin{aligned} N\Phi_2 &= NBA\cos60^\circ \\ &= 50\times0.4\times0.01\times0.5 \\ &= 0.1\ \mathrm{Wb} \end{aligned} \)
\( \begin{aligned} |\Delta(N\Phi)| &= 0.2-0.1 \\ &= 0.1\ \mathrm{Wb} \end{aligned} \)
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. …”