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Solve the following.
A circular coil of 20 turns and a radius of is placed in a uniform magnetic field of normal to the…
A circular coil of 20 turns and a radius of $10\text{ cm}$ is placed in a uniform magnetic field of $0.10\text{ T}$ normal to the plane of the coil. If the current in the coil is $5\text{ A}$, what will be:
average force on each electron in the coil due to the magnetic field. (The coil is made of copper wire of a cross-sectional area of $10^{-5}\text{ m}^2$, and the free electron density in copper is given to be about $10^{29}\text{ m}^{-3}$).
Answer
Answer
Official answer key$F = ev_dB = e\frac{I}{neA}B = 5 \times 10^{-25}$ N
Final answer: $5 \times 10^{-25}$ N
From ISC 2025 Practice Physics, question 71(c).