Current amplification α = 0.99
Change in emitter current ΔI = 5 mA
∵ α = \(\frac{\Delta I_{C}}{\Delta I_{E}}\)
∴ ΔIC = α • ΔIE = 0.99 × 5 = 4.95 mA
∵ ΔIE = ΔIE + ΔIC
=> ΔIB – ΔIE – ΔIC
= 5 mA – 4.95 mA
= 0.05 mA
∴ ΔIB = 5 × 10-5 A= 50 × 10-6 A = 50μA