Question
A toroid having average diameter $2.5\,m ,$ number $A$ turns $400,$ current $=2 \,A$ and magnetic field has $10, T$ what will be induced magnetic field (in $amp/m)$
✓
Answer
The expression to calculate the induced magnetic field is given as,
$B =\mu_{0}( H + I )$
$=\mu_{0}\left(\frac{ N }{2 \pi r } \times i + I \right)$
Substitute the values.
$10=\mu_{0}\left(\frac{ N }{2 \pi r } \times i + I \right)$
$I=\frac{10}{4 \pi \times 10^{-7}}-\frac{400 \times 2 \times 2}{2 \pi \times 2.5}$
$=\frac{10^{8}}{4 \pi}$
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