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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 `xy`-plane and a steady current I flows through the wire. The`z`-component of the magetic field at the centre of the spiral is A. `(mu_0NI)/(2(b-a))ln(b/a)`B. `(mu_0NI)/(2(b-a))ln ((b+a)/(b-a))`C. `(m_0NI)/(2b)ln(b/a)`D. `(m_0NI)/(2b)((b+a)/(b-a))` |
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Answer» Correct Answer - A a. If we take a small strip of `dr` at distance `r` from centre, then number of turns in this strip would be `dN=(N/(b-a))dr` Magnetic field due to this element at the cenntre of the coil will be `dB=(mu_)(dN)I)/(2r)=(mu_0NI)/((b-a)) (dr)/(2r)` `:. Bint_(r=a)^(r=b)dB=(m_0NI)/(2(b-a))ln(b/a)` `:. current answer is a. |
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