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A tangent galvanometer has a coil with $50$ $turns$ and radius equal to $4$ $cm$. A current of $0.1 $ $A$ is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is $7 \times {10^{ - 5}}$ $tesla$ and ${\mu _0} = 4\pi \times {10^{ - 7}}\, weber/amp \times m$, then the deflection in the galvanometer needle will be.....$^o$
$45$
$48.2$
$50.7$
$52.7$
Solution
(b)For tangent galvanometer $I = \frac{{2rB}}{{{\mu _0}n}}\tan \theta $
$\therefore \;\tan \theta = \frac{{I{\mu _0}n}}{{2rB}} = \frac{{0.1 \times 4\pi \times {{10}^{ – 7}} \times 50}}{{0.04 \times 7 \times {{10}^{ – 5}} \times 2}} = 1.12$
or $\theta = {\tan ^{ – 1}}\left( {1.12} \right) = {48.2^o}$
