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NEET PHYSICSEasy

A closely packed coil having 1000 turns has an average radius of 62.8 cm. If the current carried by the wire of the coil is 1 A, the value of the magnetic field produced at the centre of the coil will be nearly: (permeability of free space = 4π×10−74\pi \times 10^{-7} H/m)

A

10−1 T10^{-1} \text{ T}

B

10−2 T10^{-2} \text{ T}

C

102 T10^2 \text{ T}

D

10−3 T10^{-3} \text{ T}

Step-by-Step Solution

  1. Identify Formula: The magnitude of the magnetic field (BB) at the center of a circular coil with NN turns, radius RR, and carrying current II is given by the formula: B=μ0NI2RB = \frac{\mu_0 N I}{2R}
  2. Identify Parameters: Number of turns (NN): 1000=1031000 = 10^3 Current (II): 1 A1 \text{ A} Radius (RR): 62.8 cm=62.8×10−2 m62.8 \text{ cm} = 62.8 \times 10^{-2} \text{ m} Permeability (μ0\mu_0): 4π×10−7 T m/A4\pi \times 10^{-7} \text{ T m/A}
  3. Calculation: Substitute the values into the formula: B=(4π×10−7)×103×12×(62.8×10−2)B = \frac{(4\pi \times 10^{-7}) \times 10^3 \times 1}{2 \times (62.8 \times 10^{-2})} Recognizing that 2π≈6.282\pi \approx 6.28, we can write 62.8=10×(2π)62.8 = 10 \times (2\pi). B=4π×10−42×10×2π×10−2B = \frac{4\pi \times 10^{-4}}{2 \times 10 \times 2\pi \times 10^{-2}} B=4π×10−440π×10−2B = \frac{4\pi \times 10^{-4}}{40\pi \times 10^{-2}} B=110×10−2=10−1×10−2=10−3 TB = \frac{1}{10} \times 10^{-2} = 10^{-1} \times 10^{-2} = 10^{-3} \text{ T}
  4. Conclusion: The magnetic field is 10−3 T10^{-3} \text{ T}.
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