$A$ horizontal overhead powerline is at a height of $4\ m$ from the ground and carries a current of $100\ A$ from east to west. The magnetic field directly below it on the ground is: (Given: $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$)
- A$2.5 \times 10^{-7} \, T$ Northward
- B$2.5 \times 10^{-7} \, T$ Southward
- C$5 \times 10^{-6} \, T$ Northward
- D$5 \times 10^{-6} \, T$ Southward
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The magnetic field at the centre $O$ in the given figure is
MediumKCET 2019
View SolutionAs shown in the figure,two infinitely long straight parallel wires $P$ and $Q$ carrying equal currents in opposite directions are arranged parallel to the $Y$-axis. If the magnetic field due to wire $P$ at the origin '$O$' of the coordinate system is $B$,then match the resultant magnetic fields at various points given in Column $A$ with the points given in Column $B$. Column $A$ Column $B$ $A) \frac{B}{4}$ $i) (0, 0)$ $B) \frac{B}{2}$ $ii) (a, 0)$ $C) \frac{2B}{3}$ $iii) (2a, 0)$ $D) 2B$ $iv) (3a, 0)$
| Column $A$ | Column $B$ |
| $A) \frac{B}{4}$ | $i) (0, 0)$ |
| $B) \frac{B}{2}$ | $ii) (a, 0)$ |
| $C) \frac{2B}{3}$ | $iii) (2a, 0)$ |
| $D) 2B$ | $iv) (3a, 0)$ |
$A$ small cube of side $1 \ mm$ is placed at the center of a circular loop of radius $20 \ cm$. If the current in the loop is $2 \ A$,then the magnetic energy stored inside the cube is (Assume $\mu_0 = 4 \pi \times 10^{-7} \ SI$ units).
Two long straight parallel conductors $10 \text{ cm}$ apart,carry equal currents of magnitude $3 \text{ A}$ in the same direction. Then,the magnetic induction at a point midway between them is
The magnetic field at a perpendicular distance of $1 \,m$ from a long straight wire carrying a current of $1 \,A$ is:
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