CBSE-XI-Physics
35: Magnetic Field due to a Current
- #3A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field
B →due to this current.Ans : Given:
Magnitude of current, I = 10 A
Diameter of the wire, d = 1.6 × 10-3 m
∴ Radius of the wire = 0.8 × 10-3 m
The magnetic field intensity is given by
`` \,\mathrm{\,B\,}=\frac{{\mu }_{0}i}{2\,\mathrm{\,\pi \,}r}``
`` =\frac{2\times {10}^{-7}\times 20}{0.8\times {10}^{-3}}=5\,\mathrm{\,mT\,}``
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