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    Question

    In a long solenoid of 200 turns/cm carrying a current

    i, the magnetic field at center is 6.28×10-2 T. If a second solenoid with 100 turns/cm carries current i/3, its field at center is:
    A 6.28 × 10⁻² T Correct Answer Incorrect Answer
    B 1.05 × 10⁻² T Correct Answer Incorrect Answer
    C 2.09 × 10⁻² T Correct Answer Incorrect Answer
    D 3.14 × 10⁻² T Correct Answer Incorrect Answer

    Solution

    The magnetic field at the center of a long solenoid is given by the formula: B=μ0​ni where: ·        B is the magnetic field strength ·        μ0 is the permeability of free space (4π×10-7T ⋅ m/A) ·        n is the number of turns per unit length ·        i is the current flowing through the solenoid   For the first solenoid:
    Number of turns per cm, n1=200turns/cm=200×100turns/m=20000turns/m Current, i1​=i Magnetic field at the center, B1​=6.28×10-2T Using the formula: B1​= μ0​n i1​ 6.28×10-2= μ0​(20000)i (Equation 1) For the second solenoid: Number of turns per cm, n2​=100turns/cm=100×100turns/m=10000turns/m Current, i1​=i/3 Magnetic field at the center, B2​ (to be found) Using the formula: B2​= μ0n2​i2​ B2​ = μ0​(10000)(i/3) (Equation 2) To find B2​, we can divide Equation 2 by Equation 1:

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