Question

A coil of area \(3\times10^{-4}\,\mathrm{m^2}\) and \(80\) turns of a moving coil galvanometer is suspended in a uniform radial magnetic field of \(20\,\mathrm{mT}\). If the resistance of the galvanometer is \(50\,\Omega\) and its voltage sensitivity is \(200\,\mathrm{rad\,V^{-1}}\), then the torsional constant of the spring of the galvanometer is (in \(10^{-8}\,\mathrm{N\,m\,rad^{-1}}\))

• A. \(3.6\)
• B. \(4.8\)
• C. \(2.4\)
• D. \(1.2\)

Hint:

For a moving coil galvanometer, \[ \boxed{ S_V=\frac{\theta}{V} =\frac{NBA}{kR} } \] where \(k\) is the torsional constant of the spring.

Answer 1

The Correct Option is B

Step 1: Use the expression for voltage sensitivity. Voltage sensitivity is \[ S_V=\frac{\theta}{V} =\frac{NBA}{kR}, \] where \[ N=80,\qquad B=20\times10^{-3}\,\text{T}, \] \[ A=3\times10^{-4}\,\text{m}^2,\qquad R=50\,\Omega, \] and \[ S_V=200\,\text{rad V}^{-1}. \]

Step 2: Calculate the torsional constant. \[ k = \frac{NBA}{RS_V}. \] Substituting, \[ k = \frac{80\times20\times10^{-3}\times3\times10^{-4}} {50\times200} = 4.8\times10^{-8}\,\text{N\,m\,rad}^{-1}. \] Hence, \[ \boxed{4.8\times10^{-8}\,\text{N\,m\,rad}^{-1}} \] Therefore, in units of \[ 10^{-8}\,\text{N\,m\,rad}^{-1}, \] the answer is \[ \boxed{4.8} \] Thus, \[ \boxed{(B)} \] is the correct answer.