A wire loop formed joining two semi – circular wires of radii a1 and a2 carries a current i as shown in the adjoining figure. The magnietic induction at the centre O is

The magnetic field induction at the centre O due to the inner semicircular current loop of radius a1 is B1=μoi2a1 and due to the outer loop of radius a2B2=μoi2a2(downwards ⊥ to the page)Resultant induction at O is B=B1−B2=μoi21a1−1a2a1 B2Hence the correct answer is option (a)

A wire loop formed joining two semi – circular wires of radii a1 and a2 carries a current i as shown in the adjoining figure. The magnietic induction at the centre O is

The magnetic field induction at the centre O due to the inner semicircular current loop of radius a1 is B1=μoi2a1 and due to the outer loop of radius a2B2=μoi2a2(downwards ⊥ to the page)Resultant induction at O is B=B1−B2=μoi21a1−1a2a1 B2Hence the correct answer is option (a)

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A wire loop formed joining two semi – circular wires of radii a₁ and a₂ carries a current i as shown in the adjoining figure. The magnietic induction at the centre O is

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$\frac{μ_{0} i}{2} (\frac{1}{a_{1}} + \frac{1}{a_{2}})$

$\frac{μ_{0} i}{4} (\frac{1}{a_{1}} + \frac{1}{a_{2}})$

$\frac{μ_{0} i}{4} (\frac{1}{a_{1}} - \frac{1}{a_{2}})$

$\frac{μ_{0} i}{2} (\frac{1}{a_{1}} - \frac{1}{a_{2}})$

answer is D.

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Detailed Solution

The magnetic field induction at the centre O due to the inner semicircular current loop of radius a₁ is $B_{1} = \frac{μ_{o} i}{2 a_{1}}$ and due to the outer loop of radius a₂
$B_{2} = \frac{μ_{o} i}{2 a_{2}}$ (downwards $⊥$ to the page)
Resultant induction at O is $B = B_{1} - B_{2} = \frac{μ_{o} i}{2} (\frac{1}{a_{1}} - \frac{1}{a_{2}})$
$a_{1} < a_{2} ∴ B_{1} > B_{2}$
Hence the correct answer is option (a)