Ncert Exercises & Solutions

4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

NEETprep Answer:

Current in the wire, I = 35 A

Distance of a point from the wire, r = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as:

|\vec{B}| = \frac{μ_{0}}{4 π} \frac{2 l}{r}

where, μ_{0} = Permeability of free space = 4 π \times 10^{- 7} T m A^{- 1}

|\vec{B}| = \frac{4 π \times 10^{- 7}}{4 π} \times \frac{2 \times 35}{0.2} = 3.5 \times 10^{- 5} T

Hence, the magnitude of the magnetic field at a point 20 cm from the wire is

3.5 \times 10^{- 5} T