Question

A length L of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is

• A. A quarter of its first value
• B. Unaltered
• C. Four times of its first value
• D. A half of its first value

Answer 1

Answer: (C)

Four times of its first value

Answer 2

Magnetic field at the centre of current carrying coil is given by

$B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 2 \pi N i } { r }$ ⇒ $B \propto \frac { N } { r }$ ⇒ $\frac { B _ { 1 } } { B _ { 2 } } = \frac { N _ { 1 } } { N _ { 2 } } \times \frac { r _ { 2 } } { r _ { 1 } }$ .

The following figure shows that single turn coil changes to double turn coil.

N₁ = 1 N₂ = 2

r₁ = r r₂ = r / 2

B₁ = B B₂ = ?

⇒ $\frac { B } { B _ { 2 } } = \frac { 1 } { 2 } \times \frac { r / 2 } { r } = \frac { 1 } { 4 }$ ⇒ $B _ { 2 } = 4 B$

Short trick: For such type of problems remember

$B _ { 2 } = n ^ { 2 } B _ { 1 }$