An arc with radius R=0.25m and length l=π/12 meter carries 3A current. What is the magnetic field B at the center of the arc.
Magnetic field of any arc:
B=μ04π×IR×αα=ABRB=μ04π×30.25×π3=μ0 TeslaB = \frac{μ_0}{4π} \times \frac{I}{R} \times α \\ α = \frac{AB}{R} \\ B = \frac{μ_0}{4π} \times \frac{3}{0.25} \times \frac{π}{3} = μ_0 \;TeslaB=4πμ0×RI×αα=RABB=4πμ0×0.253×3π=μ0Tesla
Answer: μ0
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