First slide

Magnetic field due to a straight current carrying wire

Question

In the given figure, net magnetic field at O will be

Difficult

A

$\frac{2 μ_{0} i}{πa} \sqrt{4 - π^{2}}$
B

$\frac{μ_{0} i}{3 πa} \sqrt{4 + π^{2}}$
C

$\frac{2 μ_{0} i}{3 {πa}^{2}} \sqrt{4 + π^{2}}$
D

$\frac{2 μ_{0} i}{3 πa} \sqrt{4 - π^{2}}$

Solution

Magnetic field at O due to
$Part (1) : B_{1} = 0$
$Part (2) : B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{πi}{(a / 2)} \otimes (along - Z -axis)$
$Part (3) : B_{3} = \frac{μ_{0}}{4 π} \cdot \frac{i}{(a / 2)} (↓) (along - Y -axis)$
$Part (4): B_{4} = \frac{μ_{0}}{4 π} \cdot \frac{πi}{(3 a / 2)} ⊙ (along + Z -axis)$
$Part (5) : B_{5} = \frac{μ_{0}}{4 π} \cdot \frac{i}{(3 a / 2)} (↓) (along - Y -axis)$
$B_{2} - B_{4} = \frac{μ_{0}}{4 π} \cdot \frac{πi}{a} (2 - \frac{2}{3}) = \frac{μ_{0} i}{3 a} \otimes$
$B_{3} + B_{5} = \frac{μ_{0}}{4 π} \cdot \frac{i}{a} (2 + \frac{2}{3}) = \frac{8 μ_{0} i}{12 πa} (↓) (along - Y -axis)$
Hence, net magnetic field
$B_{net} = \sqrt{{(B_{2} - B_{4})}^{2} + {(B_{3} + B_{5})}^{2}} = \frac{μ_{0} i}{3 πa} \sqrt{π^{2} + 4}$

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