(B) 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}{4\pi} \frac{2I}{r

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

(B) The magnetic field $B$ due to a long straight current-carrying wire at a distance $r$ is given by the formula: $B = \frac{\mu_0}{4\pi} \frac{2I}{r}$ Given: Current $I = 100\, A$ Distance $r = 4\, m$ Permeability constant $\frac{\mu_0}{4\pi} = 10^{-7}\, T\, m A^{-1}$ Substituting the values: $B = 10^{-7} \times \frac{2 \times 100}{4} = 10^{-7} \times 50 = 5 \times 10^{-6}\, T$ According to the Right-Hand Thumb Rule,if the current flows from west to east,the magnetic field lines circle the wire. Directly below the wire,the magnetic field is directed towards the south.

Class 12 Physics Moving Charges and Magnetism Biot-Savart's Law and its application

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$A$ horizontal overhead power line is at a height of $4\, m$ from the ground and carries a current of $100\, A$ from west to east. The magnetic field directly below it on the ground is $(\mu_0 = 4\pi \times 10^{-7}\, T\, m A^{-1})$.

A
$5 \times 10^{-6}\, T$ northward
B
$5 \times 10^{-6}\, T$ southward
C
$2.5 \times 10^{-7}\, T$ northward
D
$2.5 \times 10^{-7}\, T$ southward

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

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$A$ current of $1.5 \, A$ is flowing through an equilateral triangle of side $9 \, cm$. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.)

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Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is

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In a wire of radius $1 \ mm$,a steady current of $2 \ A$ uniformly distributed across the cross-section of the wire is flowing. Then the magnetic field at a point $0.25 \ mm$ from the centre of the wire is (in $\mu T$)

MediumAP EAMCET 2025

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$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

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$A$ horizontal overhead power line is at a height of $4\, m$ from the ground and carries a current of $100\, A$ from west to east. The magnetic field directly below it on the ground is $(\mu_0 = 4\pi \times 10^{-7}\, T\, m A^{-1})$.

Similar Questions

$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

$A$ current of $1.5 \, A$ is flowing through an equilateral triangle of side $9 \, cm$. The magnetic field at the centroid of the triangle is (Assume that the current is flowing in the clockwise direction.)

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is

In a wire of radius $1 \ mm$,a steady current of $2 \ A$ uniformly distributed across the cross-section of the wire is flowing. Then the magnetic field at a point $0.25 \ mm$ from the centre of the wire is (in $\mu T$)

$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

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