A horizontal overhead powerline is at a heigh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The magnetic field $B$ due to a long straight current-carrying wire at a distance $R$ is given by the formula:$B = \frac{{\mu _0} I}{2 \pi R}$Give

Vedclass · Accepted Answer

$5 	imes {10^{ - 6}}\,T$ southward

Vedclass · Answer

(C) The magnetic field $B$ due to a long straight current-carrying wire at a distance $R$ is given by the formula:$B = \frac{{\mu _0} I}{2 \pi R}$Given:$I = 100\,A$$R = 4\,m$${\mu _0} = 4\pi 	imes {10^{ - 7}}\,TmA^{-1}$Substituting the values:$B = \frac{{4\pi 	imes {{10}^{ - 7}} 	imes 100}}{{2\pi 	imes 4}}$$B = \frac{{2 	imes {{10}^{ - 7}} 	imes 100}}{4}$$B = \frac{{2 	imes {{10}^{ - 5}}}}{4} = 0.5 	imes {10^{ - 5}} = 5 	imes {10^{ - 6}}\,T$Using the Right-Hand Thumb Rule,for a current flowing from east to west,the magnetic field lines below the wire point towards the south.

$A$ 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 $\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}} \right)TmA^{-1}$.

Similar Questions

The magnetic field at the centre $C$ of the arrangement shown in the figure is:

$A$ current of $1 \, A$ flows through an infinitely long straight wire. The magnetic field produced at a point $1 \, m$ away from it is:

The magnetic field at the centre of a current-carrying circular coil of radius $R$ is $B_c$ and the magnetic field at a point on its axis at a distance $R$ from its centre is $B_a$. The value of $\frac{B_c}{B_a}$ is

$A$ and $B$ are two concentric circular conductors with center $O$,carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at $O$ due to $A$ and $B$ is $1:3$,then the value of $i_1/i_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module