Magnetic field due to currentMedium
Question
A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4\pi \times 10^{- 7}SI$ units):
Options
A.4.4 mT
B.44 T
C.44 mT
D.4.4 T
Solution
Sol. [Key Concept: Magnetic field at the center of a current coil (B = μ₀NI/2r)]
The magnitude of magnetic field due to circular coil of $N$ turns is given by
$${B_{C} = \frac{\mu_{0}iN}{2R} }{= \frac{4\pi \times 10^{- 7} \times 7 \times 100}{2 \times 0.1} }{= 4.4 \times 10^{- 3}\text{ }T }{= 4.4mT }$$
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