Question

A 28-turn circular coil of radius 4.40 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = 0.010 0t + 0.040 0t2, where B is in teslas and t is in seconds. Calculate the induced emf in the coil at t = 4.20 s.

Answer 1

Answer:

ε = -0.0589V = -58.9mV

Explanation:

In order to calculate the induced emf in the coil, you use the following formula:

$\epsilon=-N\frac{d\Phi_B}{dt}=-N\frac{d(SBcos\alpha)}{dt}$         (1)

ε: induced emf = ?

N: turns of the coil = 28

ФB: magnetic flux trough the coil

S: cross sectional area of the circular coil = π.r^2

r: radius of the cross sectional area of the coil = 4.40cm = 0.044m

B: magnetic field

α: angle between the direction of the magnetic field and the direction of the normal to the cross area of the coil = 0°

You take into account that the area is constant respect to the magnetic field that cross it. Only the magnetic field is changing with time. The magnetic field depends on time as follow:

$B(t)=0.010t+0.040t^2$           (2)

You replace the expression (2) into the equation (1), evaluate the derivative, and replace the values of the other parameters for t =4.20s:

$\epsilon=-NS\frac{dB}{dt}=-NS\frac{d}{dt}[0.010t+0.040t^2]\\\\\epsilon(t)=-NS(0.010+0.080t)\\\\\epsilon(t)=-(28)(\pi(0.044m)^2)(0.010T/s+0.080T/s^2(4.20s))\\\\\epsilon(t)=-0.0589V=-58.9mV$

The induced emf in the coil is -58.9mV