JEE Mains · Physics · STD 12 - 3. current electricity
A current of \(5\; {A}\) is passing through a non-linear magnesium wire of cross-section \(0.04\; {m}^{2}\). At every point the direction of current density is at an angle of \(60^{\circ}\) with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is ....\({V} / {m}\) (Resistivity of magnesium is \(\rho=44 \times 10^{-8}\, \Omega m\))
- A \(11 \times 10^{-3}\)
- B \(11 \times 10^{-5}\)
- C \(11 \times 10^{-7}\)
- D \(11 \times 10^{-2}\)
Answer & Solution
Correct Answer
(B) \(11 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(I=\vec{J} \cdot \vec{A}=J A \cos (\theta)\) \(5=J\left(\frac{4}{100}\right) \times \cos (60)\) \(J=5 \times 50=250\, {A} / {m}^{2}\) Now, \(\vec{E}=\rho \cdot \vec{J}\) \(=44 \times 10^{-8} \times 250=11 \times 10^{-5}\, {V} / {m}\)
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