The overhead horizontal power transmission line carries a current of 75 A from east to west direction. The magnetic field above it at a vertical distance of 1.2 m will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.25\times10^{-5}\,\mathrm{T} \), towards north
The magnetic field of a long straight wire at a distance \( r \) is
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \)
\( I = 75\,\mathrm{A} \)
\( r = 1.2\,\mathrm{m} \)
\( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle B = \frac{2\times10^{-7} \times 75}{1.2} = 1.25\times10^{-5}\,\mathrm{T} \)
For the direction, use the right-hand thumb rule (or \( d\vec{l}\times\hat{r} \)). Take \( \hat{\imath} \) east, \( \hat{\jmath} \) north and \( \hat{k} \) upwards. The current is along \( -\hat{\imath} \) and the point is above the wire, along \( +\hat{k} \):
\( (-\hat{\imath})\times\hat{k} = \hat{\jmath} \)
So the field above the line points towards the north.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
