First slide

Magnetic field due to a straight current carrying wire

Question

Two long wires PQR and MNP carry equal current I as shown such that QR and NP are parallel. Find the magnetic field at origin O.

Difficult

A

$\frac{μ_{0} I}{2 \sqrt{3} πa} into the page$
B

$\frac{μ_{0} I}{2 πa} out of the page$
C

$\frac{μ_{0} I}{2 \sqrt{3} πa} out of the page$
D

$\frac{μ_{0} I}{2 πa} into the page$

Solution

Due to PQ and MN, magnetic field at O will be zero.
$\begin{array}{l} r = a \sin 60^{0} = \frac{\sqrt{3}}{2} a \\ B_{QR} = \frac{μ_{0}}{4 π} \frac{I}{r} [\sin 90^{0} - \sin 30^{0}] \otimes \\ = \frac{μ_{0}}{4 π} \frac{I}{(\sqrt{3} a / 2)} \times \frac{1}{2} \otimes = \frac{μ_{0} I}{4 π \sqrt{3} a} \otimes \\ B_{NP} = \frac{μ_{0}}{4 π} \frac{I}{r} [\sin 90^{0} + \sin 30^{0}] ⊙ = \frac{3 μ_{0} I}{4 π \sqrt{3} a} ⊙ \end{array}$
Net field at O,
$B = B_{NP} - B_{QR} = \frac{μ_{0} I}{2 \sqrt{3} πa} ⊙$

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