The magnets of two suspended coil galvanomete | vedclass

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(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$,where $N$ is the number of t

Vedclass · Accepted Answer

$1: 1$

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(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$,where $N$ is the number of turns,$I$ is the current,$A$ is the area of the coil,$B$ is the magnetic field,and $	heta$ is the angle between the normal to the coil and the magnetic field.Since the current $I$,magnetic field $B$,and number of turns $N$ are the same for both galvanometers,the ratio of the torques depends only on the ratio of the areas of the coils.For the first coil (square of side $a$): $A_1 = a^2$.For the second coil (circular of radius $r = \frac{a}{\sqrt{\pi}}$): $A_2 = \pi r^2 = \pi \left( \frac{a}{\sqrt{\pi}} ight)^2 = \pi \left( \frac{a^2}{\pi} ight) = a^2$.Since $A_1 = A_2$,the torque experienced by both coils is the same.Therefore,the ratio of the torque experienced by the first coil to that experienced by the second one is $1: 1$.

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