Question
There are 50 turns $/ \mathrm{cm}$ in a long solenoid. If 4 Amp current is flowing in the solenoid, the approximate value of mag. field along its axis at an internal point and one end will be respectively.(a) $12.6 \times 10^{-3}$ Tesla; $6.3 \times 10^{-3}$ tesla(b) $12.6 \times 10^{-3}$ Tesla; $25.1 \times 10^{-3}$ tesla(c) $25.1 \times 10^{-3}$ Tesla; $12.6 \times 10^{-3}$ tesla(d) $25.1 \times 10^{-5}$ Tesla $; 6.3 \times 10^{-5}$ tesla
Step 1
Step 1: The magnetic field inside a solenoid is given by the formula $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current flowing through the solenoid. Show more…
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A long solenoid has 200 turns per $\mathrm{cm}$ and carries a current $i$. The magnetic field at its centre is $6.28 \times 10^{-2}$ $\mathrm{Wb} / \mathrm{m}^{2}$ Another long solenoid has 100 turns per $\mathrm{cm}$ and it carries a current $\frac{i}{3} .$ The value of the magnetic field at its centre is (A) $1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}$ (B) $1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}$ (C) $1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$ (D) $1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}$
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