For a 6F6 triode tube, the plate current, plate voltage, and grid...

The plate current of a triode is generally expressed as $i_p = K(V_p + \mu V_g)^n$, where $\mu$ is the amplification factor.

Comparing the given equation $i_p = 41(V_p + 7 V_g)^{1.41}$ with the standard equation, we directly get $\mu = 7$.

Alternatively, the amplification factor is defined as $\mu = -\left(\dfrac{\partial V_p}{\partial V_g}\right)_{i_p}$.

For a constant $i_p$, differentiating the given equation yields:

$0 = 41 \times 1.41 (V_p + 7 V_g)^{0.41} (dV_p + 7 dV_g)$

$dV_p + 7 dV_g = 0$

$\mu = -\dfrac{dV_p}{dV_g} = 7$