First slide

Magnetic field due to a straight current carrying wire

Question

A long straight vertical wire carries a current of 10 A flowing upwards through it at a place where the horizontal component of the earth's magnetic induction is 0.3 gauss. Then the total magnetic induction at a point 5 cm from the wire due magnetic north of the wire is

Moderate

A

0.7 gauss
B

0.5 gauss
C

0.1 gauss
D

0.4 gauss

Solution

The magnetic induction B due to a current I ampere in a long straight conductor at a distance r metre from it is given by
$B = \frac{μ_{0}}{4 π} \times (\frac{2 I}{r})$
$\begin{array}{l} Here I = 10 A, r = 5 cm = 5 \times 10^{- 2} m \\ B = 10^{- 7} \times (\frac{2 \times 10}{5 \times 10^{- 2}}) = 0 .4 gauss \end{array}$
Since the point is due to magnetic north of the conductor, the line of action of $\vec{B}$ is along east-west direction. Magnetic induction due to earth's horizontal component is 0.3 gauss in a direction from south to north.
Resultant magnetic induction B_r is given by
$B_{r} = \sqrt{(0 {.4}^{2}) + {(0 .3)}^{2}} = 0 .5 gauss$

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