Question

Answer in detail.........​

Answer 1

Question:-

A long wire bent as shown in the figure carries currect I .If the radius of semi-circular portion is a , the magnetic induction at the centre C is :-

Answer

The magnetic induction at the centre C is

$\frac{ \mu \circ \: i}{4\pi \: a} \sqrt{\pi {}^{2} + 4}$

Explanation :-

• Step 01:- Find the direction of magntic field due to each current carrying wire and semi Circle wire using right hand rule .

So, direction of magnetic field due to both straight wire is (-k) direction

and direction of magnetic field due to semi circular wire is (-i ) direction

• Step02:- Find the magnetic field due to each current carrying wire .

So,magnetic field due to straight semi infinite current carrying wire (Bw) = μ₀I/4πa

so, Magnetic field due to semi circular current carrying wire Ba= μ₀I/4a

• Step 03:- Find Bnet

So, Bnet = 2Bw + Ba

⟹ 2* {μ₀I/4πa }( -k ) + (μ₀I/4a) (-i)

⟹μ₀I/4a[(-i)+{-2/π (-k)} ]

⟹ μ₀I/4a√[1+4/π²]

∴ Bnet =$\frac{ \mu \circ \: i}{4\pi \: a} \sqrt{\pi {}^{2} + 4}$

Option (B) is correct .