The magnetic induction at the centre O of the | vedclass

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Question

(C) The magnetic field at the centre of a circular arc of radius $R$ subtending an angle $\alpha$ at the centre is given by $B = \frac{\mu_0 I \alpha}

Vedclass · Accepted Answer

$\frac{\mu_0 I \alpha}{4\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

Vedclass · Answer

(C) The magnetic field at the centre of a circular arc of radius $R$ subtending an angle $\alpha$ at the centre is given by $B = \frac{\mu_0 I \alpha}{4\pi R}$.
In the given figure, there are two circular arcs of radii $R_1$ and $R_2$ and two straight radial segments.
The magnetic field due to the straight radial segments at the centre $O$ is zero because the current element $Idl$ and the position vector $r$ are collinear ($	heta = 0^\circ$ or $180^\circ$).
The magnetic field due to the arc of radius $R_1$ is $B_1 = \frac{\mu_0 I \alpha}{4\pi R_1}$ (directed into the page).
The magnetic field due to the arc of radius $R_2$ is $B_2 = \frac{\mu_0 I \alpha}{4\pi R_2}$ (directed out of the page).
The net magnetic field at $O$ is $B = B_1 - B_2 = \frac{\mu_0 I \alpha}{4\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.

The magnetic induction at the centre $O$ of the current-carrying bent wire shown in the adjoining figure is:

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