Physics Question. Find resistance, torque and number of scale divisions.

The coil found inside a moving-coil galvanometer has dimensions of 2 cm*2 cm and 50 turns. The diameter of the wire of the coil is 0.10 mm. The strength of the radial magnetic field at each vertical side of the coil is 500 mT. The spring that provides the counter torque has a spring constant of 2*10^-5 Nm per division of the scale of the galvanometer. A voltage of 52 mV is applied to the coil.

Determine:
(a) resistance of the coil
(b) The torque that acts on the coil.
(c) The number of scale divisions turned through by the pointer from its rest position.

(p.s Given the resistivity of the material of coil=1.7*10^-8 Ohm m)

Hello aa, the resistance R = rho * length / area
Length = 50 * 8 * 10^-2 m
Area = pi * (0.05*10^-3)^2 m^2
rho - resistivity given in ohm m
Hope you would do the rest. You get R in ohm
So current V/R = 52*10^-3 / R. You will get I in ampere
Now torque tau = N B I A
N - 50, B = 0.5 T, I found already. A already known
So tau can be found
Now to know about the scale divisions we need the way in which the scale is calibrated.
Any way @ = tau / C
tau found and C is given as force constant.
So you could find by yourself.

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