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  • Net magnetic field at the centre of the circle O due to a current carrying loop as Shown in figure is, <img alt="\left(\theta<180^{\circ}\right)" src="https://entran

Net magnetic field at the centre of the circle O due to a current carrying loop as Shown in figure is, \left(\theta<180^{\circ}\right).

Option: 1

Zero


Option: 2

Perpendicular to Paper inwards


Option: 3

Perpendicular to paper outwards


Option: 4

is perpendicular to paper inward if \theta \leq 90^{\circ} and Perpendicular to paper outwards if 90^{\circ} \leqslant \theta<180^{\circ}.


Answers (1)

best_answer

B due to \operatorname{arc}\left(B_1\right)=\frac{{\mu_0i}}{4 \pi R} \cdot \theta

B due to wire  (B_2)=\frac{\mu_0i}{4 \pi R \cos \frac{\theta}{2}} \cdot 2 \sin \frac{\theta}{2} \\

                                =\frac{\mu_0 i}{4 \pi R} 2 \tan \frac{\theta}{2}

Here, B_1 is inside and B_2 is outside but

B_2 is superior so, direction is outward. (net)

 

 

Posted by

Nehul

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