Question

For a certain dynamometer ammeter the mutual inductance M varies with deflection $\theta$ (expressed in degree) as
$M=-6cos(\theta +30°)mH$

The deflecting torque produced by a direct current of 50 mA corresponding to a deflection of 60° is

• A. $25\mu N-m$
• B. $15\mu N-m$
• C. $30\mu N-m$
• D. $10\mu N-m$

Answer 1

The correct option is B $15\mu N-m$

Rate change of mutual inductance with deflection

$\frac{dM}{d\theta }=\frac{d}{d\theta }[-6\cos (\theta +30{}^{\circ }\left)\right]=6\sin (\theta +30{}^{\circ })mH$

$\frac{dM}{d\theta }atadeflectionof60°is$

${\left(\frac{dM}{d\theta }\right)}_{\theta =60{}^{\circ }}=6\sin (60{}^{\circ }+30{}^{\circ })mH$

$=6\times {10}^{-3}H/\mathrm{deg}ree$

Deflecting torque,

$T_d=I^2\frac{dM}{d\theta }={(50\times {10}^{-3})}^2\times 6\times {10}^{-3}$

$=15\times {10}^{-6}Nm=15\mu N-m$