A straight conductor carrying current splits into two parts as shown in the figure. The radius of the circular loop is . The total magnetic field at the centre of the loop is [2019]

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Solution

Q.
A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$ . The total magnetic field at the centre $P$ of the loop is [2019]

1 Zero

2 $\frac{3 μ_{0} i}{32 R}$ , outward

3 $\frac{3 μ_{0} i}{32 R}$ , inward

4 $\frac{μ_{0} i}{2 R}$ , inward

Ans.
(1)

Magnetic field due to $i_{1} = \frac{μ_{0} i_{1}}{2 R} \frac{θ_{1}}{2 π}$ (Into the plane)

Magnetic field due to $i_{2} = \frac{μ_{0} i_{2}}{2 R} \frac{θ_{2}}{2 π}$ (Out of the plane)

For parallel combination

Now, $\frac{i_{1}}{i_{2}} = \frac{ρ l_{2}}{A} \times \frac{A}{ρ l_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow \frac{i_{1}}{i_{2}} = \frac{\frac{1}{4} (2 π R)}{\frac{3}{4} (2 π R)} = \frac{1}{3}$

$\Rightarrow i_{1} = \frac{i_{2}}{3} \Rightarrow i_{2} = 3 i_{1}$

$∴$ Net magnetic field, $= \frac{μ_{0} i_{1}}{2 R} (\frac{θ_{1}}{2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{θ_{2}}{2 π})$

$= \frac{μ_{0} i_{1}}{2 R} (\frac{3 π}{2 \times 2 π}) - \frac{μ_{0} i_{2}}{2 R} (\frac{π}{2 \times 2 π})$

$= \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{i_{2}}{4}] = \frac{μ_{0}}{2 R} [\frac{3 i_{1}}{4} - \frac{3 i_{1}}{4}] = 0$

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