C

To find the time it takes for the current in a solenoid to reach half of its maximum value, we can use the formula for current in an RL circuit:
$$I(t) = I_{max} (1 - e^{-rac{R}{L}t})$$
where R is the resistance, L is the inductance, and I_{max} is the maximum current.
At half of the maximum current, we have I(t) = 0.5 * I_{max}, which gives us:
$$0.5 I_{max} = I_{max} (1 - e^{-rac{R}{L}t})$$
Dividing by I_{max} and simplifying gives:
$$0.5 = 1 - e^{-rac{R}{L}t}$$
This leads to:
$$e^{-rac{R}{L}t} = 0.5$$
Taking the natural logarithm on both sides:
$$-rac{R}{L}t = ext{ln}(0.5)$$
Which simplifies to:
$$t = -rac{L}{R} ext{ln}(0.5)$$
Substituting the values: L = 50 mH = 50 × 10^{-3} H and R = 0.025 Ω:
$$t = -rac{50 imes 10^{-3}}{0.025} ext{ln}(0.5)$$
Calculating:
$$ ext{ln}(0.5) ext{ is approximately } -0.693$$
So:
$$t ≈ -rac{50 imes 10^{-3}}{0.025} imes (-0.693)$$
$$t ≈ rac{50 imes 0.693}{0.025}$$
$$t ≈ 1.38 ext{ ms}$$
Therefore, the closest option is 1.34 ms, which corresponds to Option C.