AP EAMCET · PHYSICS · Magnetic Effects of Current
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.)
- A \(3 \times 10^{-5} \mathrm{Nm}^{-1}\)
- B \(1 \times 10^{-5} \mathrm{Nm}^{-1}\)
- C 0
- D \(1.5 \times 10^{-5} \mathrm{Nm}^{-1}\)
Answer & Solution
Correct Answer
(A) \(3 \times 10^{-5} \mathrm{Nm}^{-1}\)
Step-by-step Solution
Detailed explanation
\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…
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