The sensitivity of the galvanometer

[Asmaa Mohammad]

Homework Statement

Homework Equations

sensitivity= θ/i

The Attempt at a Solution

That's the answer in my textbook:
Sensitivity = θ/i = 60/30 = 2 deg/mA

But I wonder why it used the angle 60 (the angle between the coil face and the field lines) instead of the angle 30 (the angle between the normal to the coil and the field lines)?
The textbook defines the sensitivity of the galvanometer as:
"The scale deflection per unit current intensity passing through its coil."

I think that the scale deflection is related to the torque (in case of a current carring coil rotates in a magnetic field) and according to its formula:
τ=BiAN sinθ (where B is the magnetic flux density, i is current intensity, A is the cross section area, N is the number of turns and θ is the angle between the normal to the plane and the magnetic field lines).
I think we need to use the angle (30) to determine the sensitivity of the galvanometer, right?

 

[cnh1995]

I believe θ in sensitivity is the angular displacement of the pointer (or the moving coil).
At zero current, the angle between face of the coil and magnetic field lines is zero.

 

[Asmaa Mohammad]

I believe θ in sensitivity is the angular displacement of the pointer (or the moving coil).
At zero current, the angle between face of the coil and magnetic field lines is zero.

Then we should use the angle 60?!

 

[cnh1995]

Then we should use the angle 60?!

Yes. Because the angular displacement of the face of the coil will be equal to the angular displacement of the normal to the plane of the coil. So if the face of the coil moves through 60 degrees, the normal too moves through 60 degrees. Hence, its angle reduces to 30 degrees from initial 90 degrees.