JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
The horizontal component of earth's magnetic field at a place is \(3.5 \times 10^{-5} \mathrm{~T}\). A very long straight conductor carrying current of \(\sqrt{2} A\) in the direction from South east to North West is placed. The force per unit length experienced by the conductor is _______ \(\times 10^{-6} \mathrm{~N} / \mathrm{m}\).
- A \(35\)
- B \(15\)
- C \(74\)
- D \(64\)
Answer & Solution
Correct Answer
(A) \(35\)
Step-by-step Solution
Detailed explanation
\( B_H=3.5 \times 10^{-5} T \) \( F=i \ell B \sin \theta, \quad \mathrm{i}=\sqrt{2} A \) \( \frac{F}{\ell}=i B \sin \theta=\sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \) \( =35 \times 10^{-6} \mathrm{~N} / \mathrm{m}\)
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