Question

(11%) Problem 9: An oscillating vertically-oriented magnetic field is given by B(t) = Bosin(ot), with Bo = 0.36 T and 0 = 1.5 rad/s. 33% Part (a) What is the magnetic flux, in webers, through a horizontal circle of radius 16 cm at t= 0? A 33% Part (b) What is the time derivative of the magnetic flux, in webers per second, through this circle at t = 0? > A 33% Part (c) What is the magnitude of the electric field, in newtons per coulomb, around the circumference of the circle at t=0?

Answer 1

Given:

$B=B_{o}\sin(\omega t)$

$B_{o}=0.36T$

$\omega = 1.5 rad/s$

a) We have a horizontal circle of radius r = 16 cm = 0.16 m.

Magnetic flux through a closed loop is given by $\phi = B*A$ [where A is the area of loop, and the plane of loop is perpendicular to magnetic field]

As the magnetic field is vertically oriented and circular loop is in horizontal plane, the magnetic flux through the loop is

$\phi = B_{o}\sin (\omega t)*\pi*r^{2}$------(i)

at t = 0, magnetic flux is zero, since $sin(0)=0$ . [answer]

b) The time derivative of magnetic flux from equation (i) is:

$\frac{d\phi}{dt} = \frac{d\left (B_{o}\sin (\omega t)*\pi*r^{2} \right )}{dt} = \pi \omega r^{2}B_{o}\cos(\omega t)$

at t=0, the time derivative of magnetic flux is $\frac{d\phi}{dt}_{t=0} = \pi \omega r^{2}B_{o}$ since $cos(0)=1$ . [answer]

c) Let the induced electric field be E.

The magnitude of e.m.f induced is $\varepsilon =\left | \frac{d\phi}{dt} \right |=\pi \omega r^{2}B_{o}\cos(\omega t)$ .

Also, the e.m.f can be expressed in terms of electric field along the circumference.

$\varepsilon =E*2\pi r$

Therefore, $E*2\pi r=\pi \omega r^{2}B_{o}\cos(\omega t)$

$\Rightarrow E=\frac{\omega rB_{o}\cos(\omega t)}{2}$

At t=0, the electric field is $E=\frac{\omega rB_{o}}{2}$ , since $cos(0)=1$ . [answer]