Question:
Figure shows a straight wire length $l$ carrying current $i$. The magnitude of magnetic field produced by the current at point $P$ is.
Figure shows a straight wire length $l$ carrying current $i$. The magnitude of magnetic field produced by the current at point $P$ is.
Updated On: Apr 26, 2024
- $\frac{\sqrt{2\mu_{0}i}}{\pi1}$
- $\frac{\sqrt{\mu_{0}i}}{4\pi1}$
- $\frac{\sqrt{2}\mu_{0}i}{8\pi1}$
- $\frac{\mu_{0}i}{2\sqrt{2}\pi1}$
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The Correct Option is C
Solution and Explanation
The given situation can be redrawn as follows. As we know the general formula for finding the magnetic field due to a finite length wire
$B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)$
Here, $\phi_{1}=0^{\circ}, \phi=45^{\circ}$
$\therefore B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$
$=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{\sqrt{2} L}$
$\Rightarrow B=\frac{\sqrt{2} \mu_{0} i}{8 \pi l}$
$B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)$
Here, $\phi_{1}=0^{\circ}, \phi=45^{\circ}$
$\therefore B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin 0^{\circ}+\sin 45^{\circ}\right)$
$=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{\sqrt{2} L}$
$\Rightarrow B=\frac{\sqrt{2} \mu_{0} i}{8 \pi l}$
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Magnets are used in many devices like electric bells, telephones, radio, loudspeakers, motors, fans, screwdrivers, lifting heavy iron loads, super-fast trains, especially in foreign countries, refrigerators, etc.
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Some of the properties of the magnetic field lines are:
- The lines and continuous and outside the magnet, the field lines originate from the North pole and terminate at the South pole
- They form closed loops traversing inside the magnet.
- But here the lines seem to originate from the South pole and terminate at the North pole to form closed loops.
- More number of close lines indicate a stronger magnetic field
- The lines do not intersect each other
- The tangent drawn at the field line gives the direction of the field at that point.