$A$ horizontal overhead power line carries a current of $90 \;A$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 \;m$ below the line?
(A) Current in the power line,$I = 90 \;A$.
Point is located below the power line at a distance,$r = 1.5 \;m$.
The magnetic field $B$ due to a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5} = 1.2 \times 10^{-5} \;T$.
The current is flowing from East to West. According to the Right-Hand Thumb Rule,if you point your thumb in the direction of the current (West),your fingers curl around the wire. Below the wire,the fingers point towards the South. Thus,the direction of the magnetic field is towards the South.
Point is located below the power line at a distance,$r = 1.5 \;m$.
The magnetic field $B$ due to a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5} = 1.2 \times 10^{-5} \;T$.
The current is flowing from East to West. According to the Right-Hand Thumb Rule,if you point your thumb in the direction of the current (West),your fingers curl around the wire. Below the wire,the fingers point towards the South. Thus,the direction of the magnetic field is towards the South.
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Two long parallel wires are at a distance $2d$ apart. They carry steady equal currents flowing out of the plane of the paper,as shown. The variation of the magnetic field $B$ along the line $XX'$ is given by
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View Solution$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?
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View Solution$A$ current of $5 \; A$ flows through a square loop of side length $\frac{1}{\sqrt{2}} \; m$. The magnitude of the magnetic field $B$ at the centre of the square loop is $p \times 10^{-6} \; T$. Find the value of $p$. [Take $\mu_0 = 4 \pi \times 10^{-7} \; T \cdot m \cdot A^{-1}$].
The magnetic field at point $O$ for the given circuits is provided. Which of the following is correct?
$(i)$ $(ii)$ $(iii)$ $(A). \frac{\mu_0 i}{2r} \odot$ $(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$ $(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$ $(B). \frac{\mu_0 i}{2r} \otimes$ $(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$ $(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$ $(C). \frac{3\mu_0 i}{8r} \otimes$ $(C). \frac{\mu_0 i}{4r} \otimes$ $(C). \text{Zero}$ $(D). \frac{3\mu_0 i}{8r} \odot$ $(D). \frac{\mu_0 i}{4r} \odot$ $(D). \text{Infinite}$
| $(i)$ | $(ii)$ | $(iii)$ |
|---|---|---|
| $(A). \frac{\mu_0 i}{2r} \odot$ | $(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$ | $(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$ |
| $(B). \frac{\mu_0 i}{2r} \otimes$ | $(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$ | $(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$ |
| $(C). \frac{3\mu_0 i}{8r} \otimes$ | $(C). \frac{\mu_0 i}{4r} \otimes$ | $(C). \text{Zero}$ |
| $(D). \frac{3\mu_0 i}{8r} \odot$ | $(D). \frac{\mu_0 i}{4r} \odot$ | $(D). \text{Infinite}$ |
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