JEE Main · 2016 · 03 Apr 2016 · Offline · Magnetic Effects of Current

Two identical wires A and B, each of length '

ℓ

', carry the same current I. Wire A is bent into a circle of radius

R

and wire B is bent to form a square of side 'a'. If

B_{A}

and

B_{B}

are the values of magnetic field at the centres of the circle and square respectively, then the ratio

\frac{B _{A}}{B _{B}}

is :

Answer: (C)

Solution

Both wires have length

ℓ

. Bent into a circle,

2 π R = ℓ

, so

R = \frac{ℓ}{2 π}

and the centre field is

B_{A} = \frac{μ _{0} I}{2 R} = \frac{μ _{0} I π}{ℓ}

. Bent into a square, each side is

a = \frac{ℓ}{4}

; one side, seen from the centre at perpendicular distance

\frac{a}{2}

, contributes

\frac{μ _{0} I}{4 π ( a /2 )} (sin 4 5^{\circ} + sin 4 5^{\circ}) = \frac{μ _{0} I 2}{2 π a}

, and the four sides give

B_{B} = \frac{4 μ _{0} I 2}{2 π a} = \frac{8 2 μ _{0} I}{π ℓ}

. The ratio is

\frac{B _{A}}{B _{B}} = \frac{μ _{0} I π / ℓ}{8 2 μ _{0} I / ( π ℓ )} = \frac{π ^{2}}{8 2}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option