Q128 / 180 · COMEDK 2015 · Physics

A certain length of insulated wire can be bent to form either a single circular loop (case I) or a double loop of smaller radius (case II). When the same steady current is passed through the wire, the ratio of the magnetic field at the centre in (case I) to that in (case II) is

  • A
    1
  • B
    2
  • C
    $1 / 2$
  • Correct answer
    D
    $1 / 4$

Solution

Magnetic field at the centre due to current carrying circular loop having $N$ turns is given as $$ B=\frac{\mu_{0} N I}{2 r} $$ If $l$ be the length of wire, then for the first case, $\begin{aligned} 2 \pi r_{1} &=l \\ \Rightarrow \quad r_{1} &=\frac{l}{2 \pi} \\ \therefore \text { Magnetic field, } B_{1} &=\frac{\mu_{0} \cdot 1 \cdot I}{2 \cdot r_{1}}=\frac{\mu_{0} I}{2 \cdot l / 2 \pi} \\ B_{1} &=\frac{\mu_{0} I \pi}{l} \quad \text{...(i)} \end{aligned}$ For the second case, $2 \times 2 \pi r_{2}=l$ $\Rightarrow \quad r_{2}=\frac{l}{4 \pi}$ $\therefore \quad B_{2}=\frac{\mu_{0} \cdot 2 \cdot I}{2 r_{2}}=\frac{2 \mu_{0} I}{2 \cdot \frac{l}{4 \pi}}$ $\Rightarrow \quad B_{2}=\frac{4 \mu_{0} I \pi}{l}=4 \times B_{1} \quad$ [from $\left.\mathrm{Eq} .(\mathrm{i})\right]$ $\Rightarrow \quad \frac{B_{1}}{B_{2}}=\frac{1}{4}$

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

Relevant Questions