1 Magnetism
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
1) $$3.0 \times 10^{-5} \, N\, m^{-1}$$ 2) $$3.2 \times 10^{-5}\, N\, m^{-1}$$ 3) $$3.6 \times 10^{-5}\, N\, m^{-1}$$ 4) $$3.8 \times 10^{-5}\, N\, m^{-1}$$
Solution:
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
$$f=IBsin90^{\circ}=1.2\times3\times10^{-5}\times1$$
$$=3.6\times10^{-5}\,N \,m^{-1}$$
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