Practicing Success
Practicing Success
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A long wire carrying current I is bent in the form of a circle as shown in the figure.
The $\vec{B}$ at point O is: |
$\frac{\mu_0}{4 \pi} \frac{2 I}{r}(1-\pi)$ perpendicularly outward $\frac{\mu_0}{4 \pi} \frac{2 \pi I}{r}$ perpendicularly inward $\frac{\mu_0}{4 \pi} \frac{2 I}{r}(1+\pi)$ perpendicularly outward $\frac{\mu_0}{4 \pi} \frac{2 I}{r}(1+\pi)$ perpendicularly inward |
$\frac{\mu_0}{4 \pi} \frac{2 I}{r}(1+\pi)$ perpendicularly inward |
Magnetic field due to straight wire is $ B_1 = \frac{\mu_0 I}{2\pi r}$ Magnetic field due to circular wire is $ B_ 2= \frac{\mu_0 I}{2 r}$ $ B = B_1 + B_2 = \frac {\mu_0}{4\pi} \frac{2I}{r}(1+\pi)$ Perpendicularly Inward
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