Question

A flip coil consits of `N` turns of circular coils which lie in a uniform magnetic field. Plane of the coils is perpendicular to the magnetic field as shown in figure. The coil is connected to a current integrator which measures the total charge passing through it. The coil is turned through `180^(@)` about the diameter. The charge passing through the coil is

A. (a) `(NBA)/(R )`
B. (b) `(sqrt(3)NBA)/(2R )`
C. ( c) `(NBA)/(sqrt(2)R )`
D. (d) `(2NBA)/(R )`

Answer 1

Correct Answer - D
(a) Initial flux through the coil, `phi_(bi) = +NBA`
Final flux through the coil, `phi_(bf) = -NBA`
When the coil is turned through `180^(@)` its flux reverses, the angle between magnetic field and areqa vector is reversed.
`Deltaphi_(B) = phi_(Bf) - phi_(Bi) = -NBA - (NBA) = -2NBA`
`Q = (Deltaphi_(B))/(R) = (2NBA)/(R)`