A coil of radius 44cm has 7 turns. Find the strength of the magnetic field at its centre, if 5A current flows through the coil. (𝜇𝑜=4𝜋×10−7𝑇𝑚/𝐴μ o =4π×10 −7 Tm/A)10×10−5𝑇10×10 −5 T5×10−7𝑇5×10 −7 T5×10−5𝑇5×10 −5 T10×10−7𝑇10×10 −7 T
Question
A coil of radius 44cm has 7 turns. Find the strength of the magnetic field at its centre, if 5A current flows through the coil. (𝜇𝑜=4𝜋×10−7𝑇𝑚/𝐴μ o =4π×10 −7 Tm/A)10×10−5𝑇10×10 −5 T5×10−7𝑇5×10 −7 T5×10−5𝑇5×10 −5 T10×10−7𝑇10×10 −7 T
Solution
The strength of the magnetic field at the center of a coil can be calculated using Ampere's Law, which states that the magnetic field (B) is equal to the permeability of free space (μ₀) times the current (I) times the number of turns (N), all divided by 2 times the radius (r).
The formula is: B = μ₀NI / 2r
Given: μ₀ = 4π×10−7 Tm/A N = 7 turns I = 5 A r = 44 cm = 0.44 m (since 1m = 100cm)
Substituting these values into the formula, we get:
B = (4π×10−7 Tm/A * 5 A * 7) / (2 * 0.44 m)
Solving this equation will give us the strength of the magnetic field at the center of the coil.
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