A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval delta t from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is B. What is the magnitude $final of the magnetic flux through the coil before it is rotated? What is the magnitude $final of the magnetic flux through the coil after it is rotated? What is the magnitude of the average emf induced in the coil? If a symbolic answer is required, your expression must include B, A, and any other variables.

A) magnitude of the magnetic flux through the coil before it is rotated = BA

B) magnitude of the magnetic flux through the coil before after it is rotated = 0

C) magnitude of the average emf induced in the coil = N|(BA)/Δt|

Explanation:

A) If we consider a loop of wire which has an Area (A) with number of turns as (N), it is initially placed perpendicular in the earth's magnetic field. It is now rotated from this position to a position where its plane is parallel to the field.

The earth's magnetic field at the position of the loop is denoted as B.

Thus, the flux through the loop before it is rotated is given by;

Φ_Bi = BACosθ

Before rotation, angle is 0

Thus,

Φ_Bi = BACos0 = BA

B) After it is rotated, the angle is now 90°.

Thus, Φ_Bf = BACos90 = 0

C) The average magnitude of the induced EMF in the coil equals the change in the flux divided by the time of change and multiplied by the number of turns.

Thus;

magnitude of the induced EMF =

|ε_av| = N |(Φ_Bf - Φ_Bi)/Δt|

From earlier, we saw that Φ_Bf = 0 and Φ_Bi = BACos0 = BA

Thus,

|ε_av| = N|(BA)/Δt|

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1