The magnetic field at the centre of the circular coil carrying current of 4A is

\frac{8 π}{3} \times 10^{- 5} T

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\frac{8 π}{3} \times 10^{- 4} T

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2 π \times 10^{- 5} T

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2 π \times 10^{- 4} T

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Solution

The correct option is C $2 π \times 10^{- 5} T$
$B = \frac{μ_{0} i}{2 r} [\frac{θ}{2 π}] = \frac{4 π 10^{- 7} \times 4}{2 \times 3 \times 10^{- 2}} [\frac{\frac{3 π}{2}}{2 π}] = 2 π \times 10^{- 5} T$

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The magnetic field at the centre of the circular coil carrying current of 4A is

The correct option is C 2π×10−5TB=μ0i2r[θ2π]=4π10−7×42×3×10−2[3π22π]=2π×10−5T

The correct option is C $2 π \times 10^{- 5} T$
$B = \frac{μ_{0} i}{2 r} [\frac{θ}{2 π}] = \frac{4 π 10^{- 7} \times 4}{2 \times 3 \times 10^{- 2}} [\frac{\frac{3 π}{2}}{2 π}] = 2 π \times 10^{- 5} T$