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]

Question

Smart Achievers · Accepted Answer

(4)

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

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

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)

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

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

Want Us to Email you About Special Offers & Updates?

Solution

Q. As shown in the figure, a long straight conductor with a semicircular arc of radius π10m 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 =4π×10-7 NA-2). [2023]

1 6 μT

2 1 μT

3 4 μT

4 3 μT

Ans. (4) BC=μ0I4πR(π) (B at centre of circular arc) =μ0I4R=4π×10-7×34×π10=3×10-6 T=3 μT