The magnitude of the magnetic field at O (centre of the circular part) due to the current-carrying coil as shown is:

Question

The magnitude of the magnetic field at O (centre of the circular part) due to the current-carrying coil as shown is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-tcpj80xc3uc27fkj.jpg" /> (A) ((μ )0 i/4 π )((3 π /a) + (√2/b)) (B) ((μ )0 i/2 π )((3 π /2 a) + (√2/b)) (C) ((μ )0 i/2 π )((π /3 a) + (3/√2 b)) (D) ((μ )0 i/4 π )((3 π /2 a) + (√2/b))

Tardigrade Admin · Accepted Answer

Magnetic field due to the circular segment

= \frac{3}{4} . \frac{μ _{0} i}{2 a}

Due to one straight wire segment

= \frac{( μ ) _{0} i}{4 πb} (s in 4 5^{\circ} + s in 0^{\circ}) = \frac{( μ ) _{0} i}{4 2 πb}

Net field

= \frac{3 ( μ ) _{0} i}{8 a} + 2 \times \frac{( μ ) _{0} i}{4 2 πb} = \frac{( μ ) _{0} i}{4 π} (\frac{3 π}{2 a} + \frac{2}{b})

Q. The magnitude of the magnetic field at $O$ (centre of the circular part) due to the current-carrying coil as shown is:

$\frac{( μ ) _{0} i}{4 π} (\frac{3 π}{a} + \frac{2}{b})$

$\frac{( μ ) _{0} i}{2 π} (\frac{3 π}{2 a} + \frac{2}{b})$

$\frac{( μ ) _{0} i}{2 π} (\frac{π}{3 a} + \frac{3}{2 b})$

$\frac{( μ ) _{0} i}{4 π} (\frac{3 π}{2 a} + \frac{2}{b})$

Q. The magnitude of the magnetic field at O (centre of the circular part) due to the current-carrying coil as shown is:

4π(μ)0​i​(a3π​+b2​​)

2π(μ)0​i​(2a3π​+b2​​)

2π(μ)0​i​(3aπ​+2​b3​)

4π(μ)0​i​(2a3π​+b2​​)

Magnetic field due to the circular segment =43​.2aμ0​i​ Due to one straight wire segment =4πb(μ)0​i​(sin45∘+sin0∘)=42​πb(μ)0​i​ Net field =8a3(μ)0​i​+2×42​πb(μ)0​i​=4π(μ)0​i​(2a3π​+b2​​)

Q. The magnitude of the magnetic field at $O$ (centre of the circular part) due to the current-carrying coil as shown is:

$\frac{( μ ) _{0} i}{4 π} (\frac{3 π}{a} + \frac{2}{b})$

$\frac{( μ ) _{0} i}{2 π} (\frac{3 π}{2 a} + \frac{2}{b})$

$\frac{( μ ) _{0} i}{2 π} (\frac{π}{3 a} + \frac{3}{2 b})$

$\frac{( μ ) _{0} i}{4 π} (\frac{3 π}{2 a} + \frac{2}{b})$