Fig. (a) shows a toroidal solenoid whose cross-section is rectangular. Find the magnetic flux through the cross-section if the current through the winding equals $I = 1.7 A$ , the total number of turns is $N = 1000$ , the ratio of the outside diameter to the inside one is $η = 1.6$ and the height is equal to $h = 5.0 cm$ .

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$\frac{μ_{0} N I h}{π} \ln η$

$\frac{μ_{0} N I h}{2 π} \ln η$

$\frac{μ_{0} N I h}{8 π} \ln η$

None of these

answer is B.

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Detailed Solution

The magnetic field at distance $r$ from center of toroid is

$B = \frac{μ_{0} N I}{2 π r}$

We consider a strip of thickness $d r$ at distance $r$ from center $O$ . The area of considered element is

$d A = h d r$

$d ϕ = \vec{B} . d = \vec{A} = B h d r \cos θ$

$∴ ϕ = \int_{a}^{b} \frac{μ_{0} N I}{2 π r} h d r = \frac{μ_{0} N I h}{2 π} \ln \frac{b}{a}$

$∴ ϕ = \frac{μ_{0} N I h}{2 π} \ln η$ Ans.

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