A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is $6.28 \times 10^{-2}$ Weber $/\mathrm{m}^{2}$. Another long solenoid has 100 turns per cm and it carries a current $\mathrm{i} / 3$. The value of the magnetic field at its centre is

(B) $1.05 \times 10^{-2}$ Weber $/\mathrm{m}^{2}$

(A) $1.05 \times 10^{-4}$ Weber $/\mathrm{m}^{2}$

(C) $1.05 \times 10^{-5}$ Weber $/\mathrm{m}^{2}$

(D) $1.05 \times 10^{-3}$ Weber $/\mathrm{m}^{2}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Magnetic Effects of Current

A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is

6.28 \times 1 0^{- 2}

Weber

/ m^{2}

. Another long solenoid has 100 turns per cm and it carries a current

i /3

. The value of the magnetic field at its centre is

Answer: (B)

The field deep inside a long solenoid is

B = μ_{0} ni

, set by the turns per unit length

n

and the current

i

. The second solenoid has half the turn density and one-third the current, so its field is

\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

of the first. Hence

B_{2} = \frac{6.28 \times 1 0 ^{- 2}}{6} \approx 1.05 \times 1 0^{- 2} Wb/m^{2}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option