A 15 cm wire carrying a current of 2 A is placed inside a current carrying solenoid perpendicular to its axis. The magnetic force acting on the wire is \( 20 \times 10^{-2}\,\mathrm{N} \). The magnetic field inside the solenoid is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.67\,\mathrm{T} \)
The force on a current-carrying wire in a magnetic field is \( F = BIl\sin\theta \). The wire is perpendicular to the solenoid's axis, which is the direction of the field, so \( \theta = 90^\circ \) and \( \sin\theta = 1 \).
\( l = 15\,\mathrm{cm} = 0.15\,\mathrm{m} \)
\( F = 20\times10^{-2}\,\mathrm{N} = 0.2\,\mathrm{N} \)
\( \displaystyle B = \frac{F}{Il} = \frac{0.2}{2 \times 0.15} = 0.67\,\mathrm{T} \)