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Physics Question on Magnetic Field

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28 x10−210^{-2} weber/m2m^{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:

A

1.05×10−2weber/m21.05\times10^{-2} weber/m^{2}

B

1.05×10−5weber/m21.05\times10^{-5} weber/m^{2}

C

1.05×10−3weber/m21.05\times10^{-3} weber/m^{2}

D

1.05×10−4weber/m21.05\times10^{-4} weber/m^{2}

Answer

1.05×10−2weber/m21.05\times10^{-2} weber/m^{2}

Explanation

Solution

Magnetic field due to a long solenoid is given by B=μ0niB=\mu_{0} ni From given data, 6.28×10−2=μ0×200×102×i…(i)6.28\times10^{-2}=\mu_{0} \times200 \times10^{2}\times i \ldots\left(i\right) and B=μ0×100×102×(i/3)…and \, \quad B=\mu_{0} \times100\times10^{2}\times\left(i / 3\right)\ldots (ii) Solving Eqs.(i) and (ii), we get B≈1.05×10−2wb/m2B \approx1.05 \times10^{-2} wb / m^{2}