A current carrying conductor is bent into a quarter of a circle of radius R as shown. The magnetic field at the centre O is

Question

A current carrying conductor is bent into a quarter of a circle of radius R as shown. The magnetic field at the centre O is (A) (μ0 I/8 R) inwards (B) (μ0 I/8 R) outwards (C) (μ0 I/4 R) inwards (D) (μ0 I/4 R) outwards

Tardigrade Admin · Accepted Answer

Magnetic field due to straight parts at the point O will be zero as the current elements and the position vectors at that point are parallel. Magnetic field due to circular arc is given by,

\overline{B_{0}} = \frac{μ _{0} I}{4 π R} (\frac{π}{2}) \otimes = \frac{μ _{0} I}{8 R} \otimes

Q. A current carrying conductor is bent into a quarter of a circle of radius $R$ as shown. The magnetic field at the centre $O$ is

$\frac{μ _{0} I}{8 R}$ inwards

$\frac{μ _{0} I}{8 R}$ outwards

$\frac{μ _{0} I}{4 R}$ inwards

$\frac{μ _{0} I}{4 R}$ outwards

Magnetic field due to straight parts at the point O will be zero as the current elements and the position vectors at that point are parallel. Magnetic field due to circular arc is given by,
$\overline{B_{0}} = \frac{μ _{0} I}{4 π R} (\frac{π}{2}) \otimes = \frac{μ _{0} I}{8 R} \otimes$

Q. A current carrying conductor is bent into a quarter of a circle of radius R as shown. The magnetic field at the centre O is

8Rμ0​I​ inwards

8Rμ0​I​ outwards

4Rμ0​I​ inwards

4Rμ0​I​ outwards