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Class 12 › Physics › Moving Charges and Magnetism
CBSE2019Class Class 12 · Physics2 Marks · Short✅ Verified
A long solenoid has 800 turns per metre length. A current of 1.6 A flows through it. Calculate the magnetic field induction at the end of the solenoid on its axis. ($\mu_0 = 4\pi \times 10^{-7}$ T·m/A)
✅ Answer & Solution
Magnetic field at the centre of a long solenoid: $B = \mu_0 nI$
At the end of a solenoid, the field is half the centre value:
$$B_{end} = \frac{\mu_0 nI}{2} = \frac{4\pi \times 10^{-7} \times 800 \times 1.6}{2}$$
$$B_{end} = \frac{4\pi \times 10^{-7} \times 1280}{2} = 4\pi \times 10^{-7} \times 640$$
$$B_{end} = 8.04 \times 10^{-4} \text{ T} \approx 8 \times 10^{-4} \text{ T}$$
