A straight conductor of length l carrying a current I, is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is

Question

A straight conductor of length l carrying a current I, is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is (A)  (π 2I/l)× 10-7  (B)  (π I/l)× 10-7  (C)  (π I/l2)× 10-7  (D)  (π I2/l)× 10-7

Tardigrade Admin · Accepted Answer

Radius, r=(1/π) Magnetic field at the centre, B=(μ0 I/4 r) =(4 π × 10-7 × I/4 × (l)π) =(π2 I × 10-7/l)

Q. A straight conductor of length $l$ carrying a current $I,$ is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is

$\frac{π ^{2} I}{l} \times 10^{- 7}$

$\frac{π I}{l} \times 10^{- 7}$

$\frac{π I}{l ^{2}} \times 10^{- 7}$

$\frac{π I ^{2}}{l} \times 10^{- 7}$

Radius, $r = \frac{1}{π}$
Magnetic field at the centre, $B = \frac{μ _{0} I}{4 r}$
$= \frac{4 π \times 1 0 ^{- 7} \times I}{4 \times \frac{l}{π}}$
$= \frac{π ^{2} I \times 1 0 ^{- 7}}{l}$

Q. A straight conductor of length l carrying a current I, is bent in the form of a semicircle. The magnetic field (in tesla) at the centre of the semicircle is

lπ2I​×10−7

lπI​×10−7

l2πI​×10−7

lπI2​×10−7