First slide

Magnetic field due to a straight current carrying wire

Question

Find the magnetic field at the centre of the circular loop as shown in the figure, when a single wire is bent to form a circular loop and also extends to form straight sections.

Difficult

A

$\frac{μ_{0} I}{2 R} ⊙$
B

$\frac{μ_{0} I}{2 R} (1 + \frac{1}{π \sqrt{2}}) ⊙$
C

$\frac{μ_{0} I}{2 R} (1 - \frac{1}{π \sqrt{2}}) \otimes$
D

$\frac{μ_{0} I}{R} (1 - \frac{1}{π \sqrt{2}}) \otimes$

Solution

Magnetic field due to AB,
$\begin{array}{l} {\vec{B}}_{1} = \frac{μ_{0} I}{4 πR} (\sin \frac{π}{2} - \sin \frac{π}{4}) \otimes \\ = \frac{μ_{0} I}{4 πR} [1 - \frac{1}{\sqrt{2}}] \otimes \end{array}$
Magnetic field due to circular loop,
${\vec{B}}_{2} = \frac{μ_{0} I}{4 R} ⊙$
Magnetic field due to straight wire BC,
$\begin{array}{l} B_{3} = \frac{μ_{0} I}{4 πR} [\sin \frac{π}{2} + \sin \frac{π}{4}] ⊙ \\ = \frac{μ_{0} I}{4 πR} [1 + \frac{1}{\sqrt{2}}] ⊙ \\ \vec{B} = {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3} \\ = (\frac{μ_{0} I}{2 R} + \frac{2 μ_{0} I}{4 πR} - \frac{1}{\sqrt{2}}) ⊙ = \frac{μ_{0} I}{2 R} [1 + \frac{1}{\sqrt{2} π}] ⊙ \end{array}$

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