Question

A straight wire of length $\left( \pi ^ { 2 } \right)$ metre is carrying a current of 2A and the magnetic field due to it is measured at a point distant 1 cm from it. If the wire is to be bent into a circle and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would be

• A. 50 : 1
• B. 1 : 50
• C. 100 : 1
• D. 1 : 100

Answer 1

Answer: (B)

1 : 50

Answer 2

If a wire of length l is bent in the form of a circle of radius r then

$2 \pi r = l$ ⇒ $r = \frac { l } { 2 \pi }$

$r = \frac { l } { 2 \pi } = \frac { \pi ^ { 2 } } { 2 \pi } = \frac { \pi } { 2 }$

Magnetic field due to straight wire

$B _ { 1 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 i } { r } = \frac { \mu _ { 0 } } { 4 \pi } \times \frac { 2 \times 2 } { 1 \times 10 ^ { - 2 } }$ also magnetic field due to circular loop $B _ { 2 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi i } { r } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi \times 2 } { \pi / 2 }$⇒ $\frac { B _ { 2 } } { B _ { 1 } } = \frac { 1 } { 50 }$