Q154 / 160 · AP EAMCET 2004 · Physics

A wire of length $l$ is bent into a circular coil of one turn of radius $R_1$. Another wire of the same material and same area of cross-section and same lengths is bent into a circular coil of two turns of radius $R_2$. When the same current flows, through the two coils, the ratio of magnetic induction at the centres of the two coils is

  • A
    $1: 2$
  • B
    $1:1$
  • Correct answer
    C
    $1:4$
  • D
    $3:1$

Solution

Magnetic induction at the centre of a circular coil $$ \begin{array}{rlrl} & B =\frac{\mu_0}{2} \cdot \frac{n i}{R} \Rightarrow B \propto \frac{n}{R} \\ \text { Here, } & n =1, n_2=2, \\ \Rightarrow & l =2 \pi R_1=2 \times 2 \pi R_2 \\ & R_2 =\frac{R_1}{2} \\ & \quad \frac{B_1}{B_2} =\frac{n_1}{n_2} \times \frac{R_2}{R_1} \\ & = \frac{1}{2} \times \frac{R_1 / 2}{r}=\frac{1}{4} \\ & \quad B_1: B_2 =1: 4 \end{array} $$

Hence, the correct answer is option: 'C'.

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