A long solenoid has 200 turns per cm and carries a current i . The magnetic field at its centre is 6.28 × 10^-2 Wb/m^2 . Another long solenoid has 100 - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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A long solenoid has 200 turns per cm and carries a current i . The magnetic field at its centre is 6.28 \times 1 0^{- 2} Wb/m^{2} . Another long solenoid has 100 turns per cm and it carries a current i /3 . The value of the magnetic field at its centre is:

Solution & Step-by-step Explanation

Magnetic field inside a solenoid B = μ_{0} ni . \frac{B _{2}}{B _{1}} = \frac{n _{2} i _{2}}{n _{1} i _{1}} Given n_{1} = 200, i_{1} = i, B_{1} = 6.28 \times 1 0^{- 2} . n_{2} = 100, i_{2} = i /3 . \frac{B _{2}}{6.28 \times 1 0 ^{- 2}} = \frac{100 \times ( i /3 )}{200 \times i} = \frac{1}{2} \times \frac{1}{3} = 1/6 B_{2} = \frac{6.28 \times 1 0 ^{- 2}}{6} \approx 1.046 \times 1 0^{- 2} Wb/m^{2} Rounding gives 1.05 \times 1 0^{- 2} .

Try it yourself before checking the explanation above.

A long solenoid has 200 turns per cm and carries a current i . The magnetic field at its centre is 6.28 \times 1 0^{- 2} Wb/m^{2} . Another long solenoid has 100 turns per cm and it carries a current i /3 . The value of the magnetic field at its centre is:

1.05 \times 1 0^{- 4} Wb/m^{2}

1.05 \times 1 0^{- 2} Wb/m^{2}

1.05 \times 1 0^{- 5} Wb/m^{2}

1.05 \times 1 0^{- 3} Wb/m^{2}

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