First slide

Magnetic field due to a current carrying Element - Biot - Savart law

Question

A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X -axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is

Moderate

A

$\vec{B} = - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$
B

$\vec{B} = \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$
C

$\vec{B} = \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} + 2 \hat{k})$
D

$\vec{B} = - \frac{μ_{0}}{4 π} \frac{I}{R} (π \hat{i} - 2 \hat{k})$

Solution

Given situation is shown in the figure.
Parallel wires 1 and 3 are semi-infinite, so magnetic field at O due to them
${\vec{B}}_{1} = {\vec{B}}_{3} = - \frac{μ_{0} I}{4 π R} \hat{k}$
Magnetic field at O due to semi-circular arc in Y Z -plane is given by
${\vec{B}}_{2} = - \frac{μ_{0} I}{4 R} \hat{i}$
$Net magnetic field at point O is given by$
$\vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3}$
$\begin{array}{l} = - \frac{μ_{0} I}{4 π R} \hat{k} - \frac{μ_{0} I}{4 R} \hat{i} - \frac{μ_{0} I}{4 π R} \hat{k} \\ = - \frac{μ_{0} I}{4 π R} (π \hat{i} + 2 \hat{k}) \end{array}$

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