AOB is a right angled isosceles triangle with sides OA = OB = a in xy plane. It is carrying current i as shown. The magnetic induction at the point (0, 0, a)

Engineering

Physics

Magnetic Field

Question

AOB is a right angled isosceles triangle with sides OA = OB = $\sqrt{2}$ a in xy plane. It is carrying current i as shown. The magnetic induction at the point (0, 0, a)

due to current in BA is $\vec{B} = \frac{μ_{0} i}{4 \sqrt{6} π a} (\hat{i} + \hat{j} + \sqrt{2} \hat{k})$

due to current in BA ^{is $\vec{B} = - \frac{μ_{0} i}{4 \sqrt{6} π a} (\hat{i} + \hat{j} - \sqrt{2} \hat{k})$}

due to current in AOB is $\vec{B} = - \frac{μ_{0} i}{2 \sqrt{3} π a} (\hat{i} + \hat{j})$

due to current in AOB is $\vec{B} = - \frac{μ_{0} i}{2 \sqrt{6} π a} (\hat{i} + \hat{j})$

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${\vec{B}}_{1} = \frac{μ_{0} I}{4 πa} [O + \frac{\sqrt{2}}{\sqrt{3}}] (- \hat{i})$

$= \frac{μ_{0} I}{4 π 9} \frac{\sqrt{2}}{\sqrt{3}} (- \hat{i}) = \frac{μ_{0} I}{2 \sqrt{6}} πa (- \hat{i})$

${\vec{B}}_{2} = \frac{μ_{0} I}{4 πa} \frac{\sqrt{2}}{\sqrt{3}} (- \hat{j}) = \frac{μ_{1} I}{2 \sqrt{6}} (- \hat{j})$