(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by the formula: $B = \frac{\mu_0 i}{2\pi r}$.Give

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(A) The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $i$ is given by the formula: $B = \frac{\mu_0 i}{2\pi r}$.Given: $i = \pi \, A$,$B = 5 \times 10^{-5} \, Wb/m^2$.Substituting the values into the formula:$5 \times 10^{-5} = \frac{\mu_0 \times \pi}{2\pi r}$.Simplifying the expression:$5 \times 10^{-5} = \frac{\mu_0}{2r}$.Solving for $r$:$r = \frac{\mu_0}{2 \times 5 \times 10^{-5}} = \frac{\mu_0}{10 \times 10^{-5}} = \frac{\mu_0}{10^{-4}} = 10^4 \mu_0 \, m$.

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

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$A$ long straight wire carries a current of $\pi \, A$. The magnetic field due to it will be $5 \times 10^{-5} \, Wb/m^2$ at what distance from the wire? $[\mu_0 = \text{permeability of free space}]$

A
$10^4 \mu_0 \, m$
B
$\frac{10^4}{\mu_0} \, m$
C
$10^6 \mu_0 \, m$
D
$\frac{10^6}{\mu_0} \, m$

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Moving Charges and Magnetism

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Biot-Savart's Law and its application

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$A$ long straight wire carries a current of $\pi \, A$. The magnetic field due to it will be $5 \times 10^{-5} \, Wb/m^2$ at what distance from the wire? $[\mu_0 = \text{permeability of free space}]$

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The magnetic field at the origin due to a current element $i d l$ placed at a point with vector position $r$ is

For the magnetic field to be maximum due to a small element of a current-carrying conductor at a point,the angle between the element and the line joining the element to the given point must be.......$^o$

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