$\begin {array} {1 1} (a)\;0 & \quad (b)\;2 \times 10^{-7}T \\ (c)\;1.35 \times 10^{-7}T & \quad (d)\;2.5 \times 10^{-7}T \end {array}$

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Consider a circle of radius $r$ between the plates and co-axial with them, i.e, its center lies on the axis of the plates and its plane is normal to the axis. By symmetry, $B$ is tangential to the circle at every point and equal in magnitude over the circle.

So, $B.\: dl = 2 \pi rB$

Using ampere’s law, $2\: \pi rB = \mu_o \times $ current passing through the area enclosed by the circle

$= \large\frac{\mu_oI_Dr^2 }{R^2}$ for $r < R$

$= \mu_oI_D\: for\: r>R$

Since, $r = 15\: cm > R = 12cm,$

So, $B = \large\frac{\mu_oI_D}{2 \pi R}$

$= 4\pi \times 10^{-7} \times \large\frac{1}{2}$$ \times 3.14 \times 0.15 = 2 \times 10^{-7} T$

Ans : (b)

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