Answer to Question #170469 in Electric Circuits for Innocent

Question

A rectangle coil of 50turns hangs vertically in a uniform magnetic field of magnitude 0.01T,so that the plan of the coil is parallel to the field.the mean height of the coil is 5cm and it's mean width is 2cm.calculate the strength of the current that must pass through the coil inorder to deflect it 30degree if the torsional constant of the suspension is 10(-19) N/m per degree

Answer 1

We know,

$τ = NIABsin\theta$

We have ,

Torsional constant$= 10^{-19} N/m$

N = 30

I = ?

$A = 0.02\times 0.05 m$

= $0.001$ m

B = 0.01 T

$\theta = 30^\circ{C}$

Hence, putting all values in the torque equation,

$Toque = k\times \theta$

$τ = NIABsin\theta \\ 10^-{19}\times\dfrac{\pi}{6} = 50\times I\times0.001\times 0.01\times\dfrac{1}{2}$

$I = \dfrac{(3.14\times10^{-19})}{50}$

$I = 0.06\times10^{-19} A$