AP EAMCET · PHYSICS · Magnetic Effects of Current
A long curved conductor carries a current I (I is a vector). A small current element of length \(\mathbf{d} l\), on the wire induces a magnetic field at a point, away from the current element. If the position vector between the current element and the point is \(\mathbf{r}\), making an angle with current element then, the induced magnetic field density; \(\mathbf{d B}\) (vector) at the point is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 \mathbf{I d} l \times \mathbf{r}}{4 \pi r}\) (perpendicular to the current element \(\mathbf{d} l\) )
- B \(\frac{\mu_0 \mathbf{I} \times \mathbf{r} \times \mathbf{d} l}{4 \pi r^2}\) (perpendicular to the current element \(\mathbf{d} l\) )
- C \(\frac{\mu_0 \mathbf{I} \times \mathbf{d} l}{r}\) (perpendicular to the plane containing the current element and position vector \(\mathbf{r}\) )
- D \(\frac{\mu_0 \mathbf{I} \times \mathbf{d} l}{4 \pi r^2}\) (perpendicular to the plane containing current element and position vector \(\mathbf{r}\) )
Answer & Solution
Correct Answer
(B) \(\frac{\mu_0 \mathbf{I} \times \mathbf{r} \times \mathbf{d} l}{4 \pi r^2}\) (perpendicular to the current element \(\mathbf{d} l\) )
Step-by-step Solution
Detailed explanation
The magnetic field \(d B=\frac{\mu_0 \mathbf{I} \times \mathbf{r} \times \mathbf{d} l}{4 \pi r^2}\)
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