A long straight wire carries a current of 35 \mathrm{~A}. What is the magnitude of the field \mathrm{B} at a point 20 \mathrm{~cm} from the wire?
A long straight wire carries a current of 35 \mathrm{~A}. What is the magnitude of the field \mathrm{B} at a point 20 \mathrm{~cm} from the wire?

Ans: We are given the following:
Current in the wire, \mathrm{I}=35 \mathrm{~A}
Distance of a point from the wire, r=20 \mathrm{~cm}=0.2 \mathrm{~m}
Magnitude of the magnetic field at this point could be given as:

    \[\begin{array}{l} \mathrm{B}=\frac{\mu_{0}}{4 \pi} \frac{21}{r} \\ \text { Where, } \mu_{0}=\text { Permeability of free space }=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1} \\ \Rightarrow \mathrm{B}=\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2} \\ \Rightarrow \mathrm{B}=3.5 \times 10^{-5} \mathrm{~T} \end{array}\]

Hence, the magnitude of the magnetic field at a point 20 \mathrm{~cm} from the wire is found to be 3.5 \times 10^{-5} \mathrm{~T}.