IIT-JEE 2011 · Paper 2 · Physics Q23

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PhysicsSingle correct+3 / -1Question 23 of 60

A long insulated copper wire is closely wound as a spiral of ' $N$ ' turns. The spiral has inner radius ' $a$ ' and outer radius ' $b$ '. The spiral lies in the X-Y plane and a steady current ' $I$ ' flows through the wire. The Z-component of the magnetic field at the center of the spiral is

Closely wound flat spiral centred at the origin of the X-Y axes, inner radius a and outer radius b marked; current I enters at the left.

\dfrac{\mu_0NI}{2(b-a)}\ln\left(\dfrac ba\right)

\dfrac{\mu_0NI}{2(b-a)}\ln\left(\dfrac{b+a}{b-a}\right)

\dfrac{\mu_0NI}{2b}\ln\left(\dfrac ba\right)

\dfrac{\mu_0NI}{2b}\ln\left(\dfrac{b+a}{b-a}\right)

Answer and worked solution

Official answer: A

Turns per unit radius. The $N$ turns are spread evenly from $a$ to $b$ , so a ring of width $dr$ holds $dN=\dfrac{N\,dr}{b-a}$ turns.

Field at the centre. Each turn of radius $r$ contributes $\dfrac{\mu_0I}{2r}$ :
$B=\int_a^b\frac{\mu_0I}{2r}\cdot\frac{N\,dr}{b-a}=\frac{\mu_0NI}{2(b-a)}\ln\frac ba.$

(A)

Topics:Biot-Savart law

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IIT-JEE 2011 · Paper 2, Physics question 23