JEE Main · 2009 · 12 Apr 2009 · Offline · Magnetic Effects of Current

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs

B C

(radius

= b

) and

D A

(radius

= a

) of the loop are joined by two straight wires AB and CD . A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is

3 0^{\circ}

. Another straight thin wire with steady current

l_{1}

flowing out of the plane of the paper is kept at the origin.

The magnitude of the magnetic field

(B)

due to loop ABCD at the origin

(O)

Answer: (B)

Solution

The straight segments

A B

and

C D

lie along radial lines through

O

, so they produce no field at the origin; only the two arcs contribute. Arc

B C

of radius

b

subtends

3 0^{\circ} = π /6

O

, giving

\frac{μ _{0} I}{4 π b} \cdot \frac{π}{6} = \frac{μ _{0} I}{24 b}

, while arc

D A

of radius

a

gives

\frac{μ _{0} I}{24 a}

in the opposite sense. The net magnitude is therefore

\frac{μ _{0} I}{24} (\frac{1}{b} - \frac{1}{a}) = \frac{μ _{0} I ( b - a )}{24 ab}

. Note how the current

I

must appear in the expression — a field is impossible without it.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option