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QPages 277-278: Q16: HC Verma: Concepts of Physics - Volume 2 Permanent Magnets Pages 277-278: Q16 Solution: A tangent galvanometer shows a deflection of 45� when 10 mA of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the tangent galvanometer formula with the moving coil galvanometer equation, incorrectly omitting the factor of 2 in the denominator ($2r$) or misplacing the permeability constant ($\mu_0$). Always derive or recall that the field is $B = \frac{\mu_0 n i}{2r}$, leading to $\tan \theta = \frac{\mu_0 n i}{2r B_H}$.

A tangent galvanometer shows a deflection of 45� when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is

B_H = 3.6 \times 10^{-5} \text{ T}

and radius of the coil is 10 cm, find the number of turns in the coil.

Step-by-Step Verified Solution

**Answer:**

n = 573 \text{ turns}

. **Step-by-Step Explanation:** Given:

\theta = 45�

,

i = 10 \text{ mA}

,

B_H = 3.6 \times 10^{-5} \text{ T}

,

r = 0.1 \text{ m}

.

\tan \theta = \frac{\mu_0 n i}{2r B_H} \Rightarrow 1 = \frac{4\pi \times 10^{-7} \times n \times 0.01}{2 \times 0.1 \times 3.6 \times 10^{-5}}

. Solving gives

n \approx 573

[citation:12].

All Questions in this Chapter

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