Question

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 Il is

• A. 1
• B. 2
• C. 1/2
• D. 1/4

Answer 1

Answer: (D)

Magnetic field at the centre of a current carrying circular wire,
$B =\frac{\mu_{0}nI}{2r} $
Let length of the wire be L.
Case I
$ 2\pi r_{1} = L, r_{1} = \frac{L}{2\pi} \:\:\:\:\: \therefore \:\:B_1 =\frac{\mu_{0}I}{2\left(L /2\pi\right)} $
Case II
$ 2 \times (2\pi r_{2}) = L, r_{2} = \frac{L}{4\pi} \:\:\:\:\: \therefore \:\:B_2 =\frac{\mu_{0}(2I)}{2\left(L /4\pi\right)} $
Hence, $\frac{B_1}{B_2} = \frac{1}{4}$