A moving coil galvanometer has a coil with 175 turns and area 1 cm^2. It uses a torsion band of tors

Q: Why do we take $\sin \alpha = 1$ in the magnetic torque formula?

The problem states the coil is placed parallel to the magnetic field. The angle $\alpha$ in the formula $\tau = NIAB \sin \alpha$ is the angle between the magnetic field and the *area vector* (which is perpendicular to the plane of the coil). Thus, $\alpha = 90^\circ$, and $\sin 90^\circ = 1$.

Q: Why must the angle $\theta$ be converted to radians?

In rotational mechanics, the torsion constant $k$ is given in N-m/rad. To ensure dimensional consistency when calculating the restoring torque $\tau = k\theta$, the angle $\theta$ must be in radians.

Q: What exactly is the torsion constant $k$?

The torsion constant $k$ represents the stiffness of the suspension wire or band. It is the amount of torque required to twist the wire by an angle of one radian.

Q: How does the number of turns $N$ affect the galvanometer's sensitivity?

Increasing the number of turns $N$ increases the magnetic torque for a given current. This means the coil will deflect more for the same current, making the galvanometer more sensitive.

Q: What would happen if the magnetic field was not parallel to the coil's plane?

If the field wasn't parallel, the magnetic torque would depend on the angle of the coil, making the scale non-linear. Real galvanometers use radial magnetic fields to ensure the field is always parallel to the coil's plane, keeping the scale linear.

Question

A moving coil galvanometer has a coil with $175$ turns and area $1 	ext{ cm}^2$. It uses a torsion band of torsion constant $10^{-6} 	ext{ N-m/rad}$. The coil is placed in a magnetic field $B$ parallel to its plane. The coil deflects by $1^\circ$ for a current of $1 	ext{ mA}$. The value of $B$ (in tesla) is approximately

Options:

(A) $10^{-3}$

(B) $10^{-4}$

(C) $10^{-1}$

(D) $10^{-2}$

Sigma Percentile · Accepted Answer

### Unraveling the Moving Coil Galvanometer

Imagine a rectangular coil suspended delicately in a magnetic field. When a tiny current flows through it, the coil experiences a twisting force—a magnetic torque. But it doesn't spin endlessly! The suspension wire twists and fights back with a restoring torque. At equilibrium, these two torques perfectly balance each other, allowing us to measure the current based on the angle of twist.

### The Battle of Torques

Let's look at the magnetic torque first. The formula is given by:

$$	au_m = N I A B \sin \alpha$$

Here, the problem states the coil's plane is parallel to the magnetic field. This means the area vector, which is always perpendicular to the plane, makes a $90^\circ$ angle with the field. So, $\sin 90^\circ = 1$, and our magnetic torque simplifies to:

$$	au_m = N I A B$$

Now for the restoring torque from the torsion band. It acts like a rotational spring. The torque is simply the torsion constant $k$ times the angle of twist $	heta$:

$$	au_r = k 	heta$$

### Equating for Equilibrium

Equating the two torques for equilibrium, we get:

$$N I A B = k 	heta$$

Since we need to find the magnetic field $B$, let's rearrange the equation to isolate $B$:

$$B = \frac{k 	heta}{N I A}$$

### The Crucial Unit Conversions

Time to plug in the numbers! But watch out for the units.

Area is in square centimeters, so we convert it to $10^{-4} 	ext{ m}^2$. Current is in milliamps, so that's $10^{-3} 	ext{ A}$. And crucially, the angle $	heta$ must be in radians! One degree is $\frac{\pi}{180}$ radians. Let's substitute all these carefully:

$$B = \frac{10^{-6} 	imes \left(\frac{\pi}{180}ight)}{175 	imes 10^{-3} 	imes 10^{-4}}$$

### The Final Calculation

Now for the arithmetic. We can simplify the powers of ten. $10^{-6}$ divided by $10^{-7}$ gives us $10$ in the numerator.

$$B = \frac{10 	imes \pi}{175 	imes 180}$$

$$B = \frac{\pi}{175 	imes 18}$$

Using $\pi \approx \frac{22}{7}$, we multiply the terms in the denominator:

$$B = \frac{22}{7 	imes 175 	imes 18}$$

After a bit of arithmetic, we find $B$ is approximately $0.998 	imes 10^{-3} 	ext{ T}$. Rounding off our result, the magnetic field $B$ is approximately $10^{-3} 	ext{ T}$. This is a beautiful application of balancing torques in a galvanometer!

Select Answer:

Visualized Solution

Equilibrium of Moving Coil Galvanometer

Magnetic Torque

Restoring Torque

Equating Torques

Substituting Values

Calculation

Final Answer

The Sigma Insight: Moving Coil Galvanometer

Unraveling the Moving Coil Galvanometer

The Battle of Torques

Equating for Equilibrium

The Crucial Unit Conversions

The Final Calculation

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