Question

A Helmholtz galvanometer has coils of radius $\frac{11}{\sqrt{5}}cm$ and the number of turns $70$. Calculate the current through the coil which produces a deflection of ${45}^o$. Given magnetic intensity due to earth along horizontal is $H=\frac{80}{\pi }A/m$

• A. $5mA$
• B. $10mA$
• C. $15mA$
• D. $25mA$

Answer 1

The correct option is A $5mA$

$H=\frac{32\pi nk}{10r\sqrt{125}}$

$where,k=\frac{I}{\tan \theta }$

Putting all the values as:

$\frac{80}{3.14}=\frac{32\times 3.14\times 70\times I}{\tan 45\times 10\times \frac{11}{\sqrt{5}}\times 5\sqrt{5}}$

$I=5.06A$

or, I = 5A