As shown in the figure, a long straight conductor with a semicircular arc of radius is carrying current . The magnitude of the magnetic field at the center of the arc is (The permeability of the vacuum ). [2023]

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Solution

Q.
As shown in the figure, a long straight conductor with a semicircular arc of radius $\frac{π}{10} m$ is carrying current $I = 3 A$ . The magnitude of the magnetic field at the center $O$ of the arc is (The permeability of the vacuum $= 4 π \times 10^{- 7} {NA}^{- 2}$ ). [2023]

1 $6 μ T$

2 $1 μ T$

3 $4 μ T$

4 $3 μ T$

Ans.
(4)

$B_{C} = \frac{μ_{0} I}{4 π R} (π) (B at centre of circular arc)$

$= \frac{μ_{0} I}{4 R} = \frac{4 π \times 10^{- 7} \times 3}{4 \times \frac{π}{10}} = 3 \times 10^{- 6} T = 3 μ T$

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