Question

# A circular coil of wire consisting of 100 turns each of radius 9 cm carries of 0.4A. The magnitude of magnetic field at the centre of the coil is:

A
2.4×104 T
B
3.5×104 T
C
2.97×104 T
D
3×104 T

Solution

## The correct option is B $$2.97 \times 10^{-4} \ T$$Here $$N = 100$$$$R = 9 cm = 9 \times 10^{-2} m$$ and $$I = 0.4 A$$Now, $$B = \dfrac{\mu_0 NI}{2R} = \dfrac{2 \pi \times 10^{-7} \times 100 \times 0.4}{9 \times 10^{-2}}$$$$= \dfrac{2 \times 3.14 \times 0.4}{9} \times 10^{-3} = 0.279 \times 10^{-3} T = 2.79 \times 10^{-4} T$$Physics

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