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Physics Question on Magnetic Field

The inner and outer radius of a toroid core are 28 cm28 \,cm and 29 cm29\, cm respectively and around the core 37003700 turns of a wire are wounded. If the current in the wire is 10 A10\, A, then the magnetic field inside the core of the toroid is

A

2.60×10−2 T2.60 \times 10^{-2}\, T

B

2.60×10−3 T2.60 \times 10^{-3}\, T

C

4.52×10−2 T4.52 \times 10^{-2}\, T

D

4.52×10−3 T4.52 \times 10^{-3}\, T

Answer

2.60×10−2 T2.60 \times 10^{-2}\, T

Explanation

Solution

The number of turns per unit length for the given toroid n=N2πravn=\frac{N}{2\pi r_{av}} The average radius of toroid rav=28+292=28.5 cmr_{av}=\frac{28+29}{2}=28.5\,cm =28.5×10−2 m=28.5\times10^{-2}\,m ∴n=37002×3.14×28.5×10−2\therefore n=\frac{3700}{2\times3.14\times28.5\times10^{-2}} =2067.27≈2067=2067.27 \approx2067 Now, B=μ0nI=4π×10−7×2067×10B=\mu_{0}nI=4\pi\times10^{-7}\times2067\times10 =259615.2×10−7 T=259615.2\times10^{-7}\, T =2.60×10−2 T=2.60\times10^{-2}\, T