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Example · Example 3

Q.A single circular turn of wire of radius 10 cm10\ \text{cm} carries a current of 5 A5\ \text{A}. Find the magnetic field at the centre of the loop.

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✓ Free question

Given I=5 AI=5\ \text{A}, R=0.1 mR=0.1\ \text{m}:

B=μ0I2R=(4π×10−7)(5)2(0.1)=6.283×10−60.2=3.14×10−5 TB = \frac{\mu_0 I}{2R} = \frac{(4\pi\times 10^{-7})(5)}{2(0.1)} = \frac{6.283\times 10^{-6}}{0.2} = 3.14\times 10^{-5}\ \text{T}

✓Final answer

The magnetic field at the centre of the loop is B≈3.14×10−5 TB \approx 3.14\times 10^{-5}\ \text{T}.

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