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 $20 × 10^{-2} N$. The magnetic field inside the solenoid is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $0.67 T$
Magnetic force on a current carrying wire:
$F = BIL \sin\theta$
Here, the wire is perpendicular to solenoid axis ⇒ $\theta = 90^\circ$, so $\sin\theta = 1$.
Given: $F = 20 \times 10^{-2} \, N = 0.20 \, N$, $I = 2 \, A$, $L = 15 \, cm = 0.15 \, m$
$B = \frac{F}{IL}$
$B = \frac{0.20}{2 \times 0.15}$
$B = \frac{0.20}{0.30} = 0.667 \, T$
Final Answer:
$B = 0.667 \, T$