Question

The horizontal component of earths 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

• 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)

Force per unit length $f=\frac{F}{l}=IB\, sin\,\theta $
When the current is flowing from east to west then $\theta=90^{\circ}$, hence
$f=IBsin90^{\circ}=1.2\times3\times10^{-5}\times1$
$=3.6\times10^{-5}\,N \,m^{-1}$