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Q.A solenoid of length 2 meter and 100 turns carries a current 10A. Calculate the magnitude of the magnetic field inside the solenoid. (mu_0 = 4pi10^-7) OR A long straight wire carries a current of 30A. Calculate the magnitude of magnetic field at a point 30 cm from the wire.

Madhya Pradesh MpbseMP Board Higher Secondary 2024Subjective· 3mImportance★★★★★
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Solenoid: B = μ₀nI ≈ 6.28×10⁻⁴ T. (OR straight wire: B = μ₀I/2πr = 2×10⁻⁵ T.)

Magnetic field inside a solenoid:

Given: length l=2l = 2 m, number of turns N=100N = 100, current I=10I = 10 A, μ0=4π×10−7\mu_0 = 4\pi\times10^{-7} T·m/A.

Number of turns per unit length:

n=Nl=1002=50 turns/mn = \frac{N}{l} = \frac{100}{2} = 50\ \text{turns/m}

The magnetic field inside a long solenoid (well away from the ends) is:

B=μ0nI=(4π×10−7)(50)(10)B = \mu_0 n I = (4\pi\times10^{-7})(50)(10)

B=4π×10−7×500=2000π×10−7≈6.28×10−4 TB = 4\pi\times10^{-7}\times 500 = 2000\pi\times10^{-7} \approx 6.28\times10^{-4}\ \text{T}

OR — Magnetic field due to a long straight current-carrying wire:

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