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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Magnetism and Matter
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A copper wire of diameter $1.6mm$ carries a current i. The maximum magnetic field due to this wire is $5 \times 10^{-3} T$. The value of $i$ is

$\begin {array} {1 1} (a)\;40A & \quad (b)\;20A \\ (c)\;10A & \quad (d)\;5A \end {array}$

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Maximum field is at the surface and is given by
$B = \large\frac{ \mu_oi}{2\pi R}$
On substituting the values of the other variables, we get $i = 20A$
Ans : (b)
answered Feb 19, 2014 by thanvigandhi_1

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