A uniform wire of diameter $d$ carries a current of $50 \mathrm{~mA}$ when the mean drift velocity o

Question

A uniform wire of diameter $d$ carries a current of $50 \mathrm{~mA}$ when the mean drift velocity of free electrons is $v$. For another wire of diameter $d/3$ made of the exact same material to carry a current of $150 \mathrm{~mA}$, its mean drift velocity of electrons must be

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(b) : The current is related to drift velocity by $I = n A e v_d = n (\frac{\pi d^2}{4}) e v_d$. This implies $v_d \propto \frac{I}{d^2}$. For the first wire, $v \propto \frac{50}{d^2}$. For the second wire, $v' \propto \frac{150}{(d/3)^2} = \frac{150}{d^2 / 9} = \frac{1350}{d^2}$. Taking the ratio: $\frac{v'}{v} = \frac{1350}{50} = 27$. Therefore, the new drift velocity is $v' = 27v$.