Verified Solution • 2025–26 Session moderate

QPages 306-313: Q23: HC Verma: Concepts of Physics - Volume 2 Electromagnetic Induction Pages 306-313: Q23 Solution: A coil of radius 10 cm and resistance has 1000 turns. It is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the 180° turn as making the coil’s plane horizontal, assuming the flux drops to zero; however the coil rotates about a vertical diameter, so the normal reverses direction and the flux changes from $+BA$ to $-BA$, giving $\Delta\Phi =2BA$. Remember to use the coil’s area $A=\pi r^2$ and include the $N$ turns when converting the induced emf to total charge $q=N\Delta\Phi/R$.

A coil of radius 10 cm and resistance 40 Ω40 \ \Omega has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180�. Find the charge which flows through the galvanometer if the horizontal component of the earth's magnetic field is BH=3.0×10−5 TB_H = 3.0 \times 10^{-5} \text{ T}.

Step-by-Step Verified Solution

**Answer:** q = 4.71 _ 10__ C. **Step-by-Step Explanation:** _ = BA = 3.0 _ 10__ _ �(0.1)_ = 9.42 _ 10__ Wb. After 180� rotation, __ = 2_. q = N__/R = 1000 _ 2 _ 9.42 _ 10__/40 [citation:8].

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