First slide

Magnetic field due to a straight current carrying wire

Question

If a current of 5 A is passed through a straight wire of length 6 cm, then the magnetic induction at a point 5 cm from the either end of the wire is

Moderate

A

0.25 gauss
B

0.125 gauss
C

0.15 gauss
D

0.30 gauss

Solution

Let AB represents the current carrying conductor and P the point where magnetic induction B is required.
$\begin{array}{l} Now, B = \frac{μ_{0} I}{4 π} [\sin ϕ_{1} + \sin ϕ_{2}] \\ Here, r = OP \\ Now, AO = OB = \frac{6}{2} cm = 3 cm \\ And PB = PA = 5 cm \end{array}$
$\begin{array}{l} OP = \sqrt{{(PB)}^{2} - {(OB)}^{2}} \\ = \sqrt{{(5)}^{2} - {(3)}^{2}} = 4 cm \\ r = 4 cm = 4 \times 10^{- 2} metre \\ \sin ϕ_{1} = \frac{3}{5} and \sin ϕ_{2} = \frac{3}{5} \\ B = 10^{- 7} (\frac{5}{4 \times 10^{- 2}}) [\frac{3}{5} + \frac{3}{5}] \\ = \frac{6}{4} \times 10^{- 5} = 0 .15 \times 10^{- 4} tesla \\ = 0 .15 gauss \end{array}$
The direction of magnetic induction is normal to the plane of ABP.

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