(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

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

(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 \times {10^{ - 7}}\,TmA^{-1}$Substituting the values:$B = \frac{{4\pi \times {{10}^{ - 7}} \times 100}}{{2\pi \times 4}}$$B = \frac{{2 \times {{10}^{ - 7}} \times 100}}{4}$$B = \frac{{2 \times {{10}^{ - 5}}}}{4} = 0.5 \times {10^{ - 5}} = 5 \times {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.

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

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

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

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

When a certain length of wire is turned into one circular loop,the magnetic induction at the centre of the coil due to a current $I$ flowing through it is $B_1$. If the same wire is turned into three loops to make a circular coil,the magnetic induction at the center of this coil for the same current will be:

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$A$ helium nucleus makes a full rotation in a circle of radius $0.8 \ m$ in $2 \ s$. The value of the magnetic field $B$ at the centre of the circle will be

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Equal currents $I$ are flowing in three infinitely long wires along the positive $x$,$y$,and $z$ directions. The magnetic field at a point $(0, 0, -a)$ would be:

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$A$ vertical straight conductor carries a current vertically upwards. $A$ point $P$ lies to the east of it at a small distance and another point $Q$ lies to the west at the same distance. The magnetic field at $P$ is

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$A$ small cube of side $1 \text{ mm}$ is placed at the centre of a circular loop of radius $10 \text{ cm}$ carrying a current of $2 \text{ A}$. The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \text{ J}$. The value of $\alpha$ is . . . . . . . ($\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$,$\pi = 3.14$)

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

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$A$ helium nucleus makes a full rotation in a circle of radius $0.8 \ m$ in $2 \ s$. The value of the magnetic field $B$ at the centre of the circle will be

Equal currents $I$ are flowing in three infinitely long wires along the positive $x$,$y$,and $z$ directions. The magnetic field at a point $(0, 0, -a)$ would be:

$A$ vertical straight conductor carries a current vertically upwards. $A$ point $P$ lies to the east of it at a small distance and another point $Q$ lies to the west at the same distance. The magnetic field at $P$ is

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