Question

The horizontal component of earth's magnetic field at a certain place is $3.0\times {10}^{-5}$ T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of 1.2 A in east to west direction is then

• A. $3.0\times {10}^{-5}N{m}^{-1}$
• B. $3.2\times {10}^{-5}N{m}^{-1}$
• C. $3.6\times {10}^{-5}N{m}^{-1}$
• D. $3.8\times {10}^{-5}N{m}^{-1}$

Answer 1

Answer: (C)

$3.6\times {10}^{-5}N{m}^{-1}$

Force per unit length $f=\frac{F}{l}=IBsin\theta$
When the current is flowing from east to west then $\theta ={90}^o$, hence
$f=IBsin{90}^o=1.2\times 3\times {10}^{-5}\times 1=3.6\times {10}^{-5}N{m}^{-1}$