A single loop of area 0.193 m2 is placed perpendicular to a magnetic field of 0.374 T. The loop is rotated about an axis lying in the plane of the loop and passing through its center. An instantaneous emf of 9.21 Volts is generated when the plane of the loop makes an angle of 45 o with the field direction. What is the angular speed of the loop (rad/s)

The angular speed w of the loop = 181 rad/s

Explanation:

Given that;

Area A = 0.193 m2 

Magnetic field B = 0.374 T

E.m.f = 9.24v

Ø = 45 degree

According to Faraday's law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.

Using the formula

E.m.f = NABwcosØ

Where w = angular velocity.

Let assume that N = 1 then,

9.24 = 0.193 × 0.374 × w × cos45

9.24 = 0.051w

w = 9.24/0.051

w = 181 m/s

Therefore, the angular speed w of the loop = 181 rad/s

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we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

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so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft