The force acting per unit length when a very long straight conductor is carrying a steady current of $1…

The force acting per unit length when a very long straight conductor is carrying a steady current of $1 \mathrm{~A}$ and the direction of the current is from south to north is (The horizontal component of the earth's magnetic field at the place is $3 \times 10^{-5} \mathrm{~T}$ and the direction of the field is from the geographical south to geographical north.)
  1. $3 \times 10^{-5} \mathrm{Nm}^{-1}$
  2. $1 \times 10^{-5} \mathrm{Nm}^{-1}$
  3. 0
  4. $1.5 \times 10^{-5} \mathrm{Nm}^{-1}$

Solution

$\begin{aligned} & \text { Steady current, } \mathrm{i}=1 \mathrm{~A} \\ & \text { Magnetic field } \mathrm{B}=3 \times 10^{-5} \mathrm{~T} \\ & \mathrm{~F}=\mathrm{il} \times \mathrm{B}=\mathrm{ilB} \sin \theta \\ & \frac{\mathrm{F}}{1}=\mathrm{iB} \sin 90 \\ & =1 \times 3 \times 10^{-5} \times 1 \\ & =3 \times 10^{-5} \mathrm{~N} / \mathrm{m}\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 1)

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