Question

A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of the magnetic field B changes in time according to the equation B ( t ) = ( 3.75 T ) + ( 3.05 T s ) t + ( − 6.95 T s 2 ) t 2 If the radius of the wire loop is 0.220 m , find the magnitude E of the induced emf in the loop when t = 5.71 s .

Answer 1

Answer:

10.90V

Explanation:

Faraday's law of induction states that the electromotive force (E) produced in a magnetic field is directly proportional to the change in flux, ΔΦ, inversely proportional to change in time, Δt. i.e

E = - N ΔΦ / Δt       ----------------(i)

Where;

N = proportionality constant called the number of coils in the wire.

With a small change in time, equation (i) could be re-written as follows;

E = - N δΦ / δt              --------------(ii)

Also, the magnetic flux, Φ, is given as follows;

Φ = BA cos θ          --------------------(iii)

Where;

B = magnetic field

A = cross sectional area of the wire

θ = angle between the field and the cross-section of the wire

Substitute equation (iii) into equation (ii) as follows;

E = - N δ(BAcosθ) / δt

E = - NAcosθ δ(B) / δt               -----------------------(iv)

From the question;

B(t) = (3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²           -------------------(v)

Substitute equation (v) into equation (iv) as follows;

E = - NAcosθ δ[(3.75 T) + (3.05 T/s)t + (− 6.95 T/s²) t²] / δt         -----(vi)

Solve equation (vi) by taking derivative as follows;

E = - NAcosθ [3.05 − (2)6.95 t]

E = - NAcosθ [3.05 − 13.9t]            ----------------(vii)

Solve for A using the following relation;

A = πr²    -----------------(viii)

Where;

r = radius of the wire loop = 0.220m

π = 3.142

Substitute these values into equation (viii) as follows;

A = 3.142 x 0.220²

A = 0.152m²

Now substitute A = 0.152m², N = 1 (a single coil), θ = 19.5° and t = 5.71s into equation (vii)

E = - (1) (0.152)cos(19.5)° [3.05 − 13.9(5.71)]

E = - (1) (0.152)(0.94) [3.05 − 13.9(5.71)]

E = 10.90V

Therefore, the induced EMF in the loop is 10.90V

Answer 2

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

$\Phi = BA Cos \theta$

Here,

$\theta$ = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

$\epsilon= -N \frac{d\Phi }{dt}$

$\epsilon = -N \frac{(BAcos\theta)}{dt}$

$\epsilon = -NAcos\theta \frac{dB}{dt}$

$\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)$

$\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )$

At time $t = 5.71s$,  Induced emf is,

$\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)( (3.05T/s)-(13.9T/s)(5.71s))$

$\epsilon = 10.9V$

Therefore the magnitude of the induced emf is 10.9V