First slide

Magnetic field due to current element

Question

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

Moderate

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

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