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Magnetic Field Due to a Circular Ring on the Axis

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Question

Find the magnetic induction of the field at the point O of a loop with current I, whose shape is illustrated in the figure.

$\frac{μ_{0} I}{3 π} (\frac{3 π}{2 a} + \frac{\sqrt{2}}{b})$

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$\frac{μ_{0} I}{2 π} (\frac{3 π}{4 a} + \frac{\sqrt{2}}{2 b})$

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$\frac{μ_{0} I}{2 π} (\frac{2 π}{3 a} + \frac{\sqrt{2}}{b})$

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$\frac{μ_{0} I}{2 π} (\frac{3 π}{4 a} + \frac{\sqrt{2}}{2 b})$

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Solution

The correct option is B
$\frac{μ_{0} I}{2 π} (\frac{3 π}{4 a} + \frac{\sqrt{2}}{2 b})$

$B_{C} = \frac{μ_{0} I}{4 π a} (\frac{3}{2} π)$
$B_{S} = 2 \times \frac{μ_{0} I}{4 π b} (s i n 0^{\circ} + s i n 45^{\circ}) = \sqrt{2} \times \frac{μ_{0} I}{4 π b}$
Now, $B = B_{C} + B_{S}$
$= \frac{μ_{0} I}{2 π} (\frac{3 π}{4 a} + \frac{\sqrt{2}}{2 b})$

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Find the magnetic induction of the field at the point O of a loop with current I, whose shape is illustrated in the figure.

The correct option is B μ0I2π(3π4a+√22b) BC=μ0I4πa(32π) BS=2×μ0I4πb(sin 0∘+sin 45∘)=√2×μ0I4πb Now, B=BC+BS =μ0I2π(3π4a+√22b)