A 800 turn coil of effective area 0.05 m² is kept perpendicular to a magnetic field 5 × 10⁻⁵ T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :

CONCEPT:

Magnetic flux: The magnetic flux is defined as the number of magnetic fields which passes through the closed surface and it is written as;

\(\phi_B=B.A\)

⇒ \(\phi_B=BAcos\theta\)       ----(1)

Induced EMF - An emf is induced in the coil when the magnetic field is pushed in and out of the coil and it is written as;

\(EMF = - \frac{{\Delta \phi }}{{\Delta t}} \)      -----(2)

Here, \(\Delta\phi\) is the change in the magnetic field, \(\Delta t\) is the change in time.

EXPLANATION:

Given: Magnetic field B = 5 × 10^–5 T

Number of turns in coil say, N = 800

Area of coil A = 0.05 m²

and the time is taken to rotate Δ t = 0.1 s

Initial angle θ₁ = 0°

Final angle θ₂ = 90°

Change in magnetic flux \(\Delta \phi \) ,using equation 1) we have,

\(\Delta \phi \) = NBAcos90° – BAcos0°

⇒ \(\Delta \phi \) = – NBA

⇒ \(\Delta \phi \) = – 800 × 5 × 10^–5 × 0.05

⇒ \(\Delta \phi \) = – 2 × 10^–3 weber

Now, by using the equation (2) we have;

\(EMF = - \frac{{\Delta \phi }}{{\Delta t}}\)

⇒ \( EMF= \frac{{ - ( - )2 \times {{10}^{ - 3}}Wb}}{{0.1s}} \)

⇒ EMF = 0.02 V

Hence, option 4) is the correct answer.