Question

I apply an electric field of ${10}^{5}N/C$ on a conductor of cross section $4m{m}^2$ and I get a current of 1A. If I want to change the current to 0.3A, what is the electric field I need to apply?

• A. $3\times {10}^{4}N/C$
• B. $5\times {10}^{4}N/C$
• C. $4\times {10}^{4}N/C$
• D. $2\times {10}^{4}N/C$

Answer 1

The correct option is A $3\times {10}^{4}N/C$

Given $E_1={10}^{5}N/C$
$I_1=1A$
A = $4m{m}^2$
$J_1$ = $I_1/A=0.25\times {10}^{6}A/m^2$

We need I = 0.3 A
So $J_2=0.75\times {10}^{6}A/m^2$

We know that
$E\alpha J$
So, $\frac{E_1}{J_1}=\frac{E_2}{J_2}$

$E_2=3\times {10}^{4}N/C$