A current I flows in a straight wire bent in the shape of a quarter ring of radius R as shown in figure. The magnetic field induction (B) at point O is

Question

A current I flows in a straight wire bent in the shape of a quarter ring of radius R as shown in figure. The magnetic field induction (B) at point O is (A) (μ0 I/4 R) (B) (μ0 I/8 R) (C) (μ0 I/2 R) (D) (μ0 I/6 R)

Tardigrade Admin · Accepted Answer

Magnetic field at centre of a complete current carrying circular loop is

B = \frac{μ _{0} I}{2 R}

For one quarter of loop,

B = \frac{1}{4} \times \frac{μ _{0} I}{2 R} = \frac{1}{8} (\frac{μ _{0} I}{R}) T

Q. A current I flows in a straight wire bent in the shape of a quarter ring of radius $R$ as shown in figure. The magnetic field induction $(B)$ at point $O$ is

$\frac{μ _{0} I}{4 R}$

$\frac{μ _{0} I}{8 R}$

$\frac{μ _{0} I}{2 R}$

$\frac{μ _{0} I}{6 R}$

Magnetic field at centre of a complete current carrying circular loop is
$B = \frac{μ _{0} I}{2 R}$

For one quarter of loop,
$B = \frac{1}{4} \times \frac{μ _{0} I}{2 R} = \frac{1}{8} (\frac{μ _{0} I}{R}) T$

Q. A current I flows in a straight wire bent in the shape of a quarter ring of radius R as shown in figure. The magnetic field induction (B) at point O is

4Rμ0​I​

8Rμ0​I​

2Rμ0​I​

6Rμ0​I​