Question

For the cirrent carrying loop shown, the net magnetic field at the center of the circle $O$ due to it, for $\theta <{180}^o$ is $B$. Then $B$ is.

• A. Perpendicular to paper inwards
• B. Perpendicular to paper outwards
• C. Perpendicular to paper inwards if $\theta \le {90}^o$ and perpendicular to paper outwards if ${90}^o\le \theta \le {180}^o$
• D. None of them

Answer 1

Answer: (A), (B)

The correct options are
A Perpendicular to paper inwards
B Perpendicular to paper outwards

For this figure circuit carrying loop shown, the net magnetic field at the centre of the circle $O$ due to it, for

$\theta <180°$ is B, then B is

Calculation of in the field due to the two segments and add them up. The direction of field due to the circular part of the loop in perpendicular to the paper inwards and due to the straight portion of loop is outward.

They will be in opposite directions and the field due to the straight wire dominates as it in nearer the centre $O$.