First slide

Magnetic field due to a straight current carrying wire

Question

A long wire carries a current of 20 A along the axis of a solenoid. The field due to the solenoid is 4 mT. The resultant field at a point 3 mm from the solenoid axis is

Moderate

A

1.33mT
B

4.2 mT
C

2.1mT
D

8.4mT

Solution

The fielddue to wire is given by
$\begin{array}{l} B_{wire} = \frac{μ_{0}}{4 π} \times \frac{2 i}{r} tesla \\ = \frac{10^{- 7} \times 2 \times 20}{3 \times 10^{- 3}} = 1 .33 \times 10^{- 3} T \\ = 4 / 3 mT \end{array}$
Field due to solenoid
B_sol = 4mT
Resultant magnetic field $= \sqrt{{(B_{wire})}^{2} + {(B_{sol})}^{2}}$
$\begin{array}{l} = \sqrt{(4 / 3)^{2} + (4)^{2}} mT \\ = 4 \cdot 2 mT \end{array}$

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