Give $\mathrm{SI}$ unit of magnetic field from Biot-Savart law.
Give $\mathrm{SI}$ unit of magnetic field from Biot-Savart law.
According to Biot-Savart law,
$d \mathrm{~B} =\frac{\mu_{0}}{4 \pi} \mathrm{I} d l \frac{\sin \theta}{r^{2}}$
$d \mathrm{~B} =\left(\frac{\mathrm{T} \cdot \mathrm{m}}{\mathrm{A}}\right)\left(\frac{\mathrm{A} \cdot \mathrm{m}}{\mathrm{m}^{2}}\right)$
$=\mathrm{T} \text { (Tesla) }$
If we take $\mathrm{I}=1 \mathrm{~A}, d l=1 \mathrm{~m}, r=1 \mathrm{~m}$ and $\theta=90^{\circ}$, then $d \mathrm{~B}=\frac{\mu_{0}}{4 \pi}=\frac{4 \pi \times 10^{-7}}{4 \pi}=10^{-7}$ tesla $\therefore 1$ tesla $=10^{7} d \mathrm{~B}$
Thus, one tesla is $10^{7}$ times the magnetic field produced by a conducting wire of length one meter and carrying current of one ampere at a distance of one meter from it and perpendicular to it.
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