Question

The magnetic field in the region between the poles of an electromagnet is uniform at any time, butis a point) increasing at the rate of 0.020 T/s. The area of the conducting loop in the field is 180 cm?. What is the magnitude of the induced emf? 0.12 mV 0.90 ml 0.11 mV 0:0.36 mV A rectangular loop with area A rotates with constant angular velocity o about the axis shown. How does (point) e sign of e vary with each rotation?

can you please show me how to do each of these thank you!

Answer 1

(1)

............ans(D)

as per faraday's law .........

emf do dt

where .......

$\phi= \overset{\rightarrow}{A} . \overset{\rightarrow}{B} = AB cos\theta$ .............................(1)

A is the area vector

B is magnetic fiels vector

$\theta$ is the angle betweem area vector and magnetic field vector.

A.BABcosoABcos0AB

as $\theta =0^{\circ}$

d(AB) dt emf =

emf = A dt

putting given values.......

4 = 180crn 2-0.01877-

$\frac{\mathrm{d} B}{\mathrm{d} t} = 0.020 Ts^{-1}$

we finds..

= 0.36 mV

emf 3.6 x 104VV

(2)  

.......... ans(B)

as per faradays law...

emf do dt

$\phi= \overset{\rightarrow}{A} . \overset{\rightarrow}{B} = AB cos\theta$

so..........

d(ABcost) emf-

$\Rightarrow emf = AB \frac{\mathrm{d} (cos\theta)}{\mathrm{d} t}$..............................[as A,B are kept constant]

$\theta = \omega t + \omega _0$

here.....  $\omega _0$ = initial angle between area vector and magnetic field vector

em-ABcosut+) dt

$\Rightarrow emf = -AB\omega sin(\omega t + \omega_0)$

$\Rightarrow emf = -AB\omega sin(\theta)$

$\theta$= angle between field and area vector

in one rotation $\theta$ varies as.... $90^{\circ} \rightarrow 180^{\circ} \rightarrow 270^{\circ} \rightarrow 360^{\circ} \rightarrow 90^{\circ}$

sign of emf depends on $sin(\theta)$ only as other factors are constant........

as .......... $sin(\theta)$ for is >0

θ 90 180

and $sin(\theta)$ for $\theta = 180^{\circ} \rightarrow 360 ^{\circ}$ <0

and $sin(\theta)$ for $\theta = 360^{\circ} \rightarrow 90 ^{\circ}$ >0

which means sign changes 2 times for the emf in the whole proscess.

(3)

.................ans(C)

The actual emf induced in the armature coil is alternating in nature. and hence , the current in the armature coil is also alternating in nature. Commutator reverses current direction at on instant when the armature coil crosses the magnetic neutral axis. So as output we gets a direct current.

(4)

As per faradays law........

emf do dt

area for one turn of coil = $\pi r^2$

as r = 2.50 cm = 0.025 m

$\pi r^2$ = 0.0019634954 $m^2$

for 500 circular turns  

$emf = {500}\frac{d(AB) }{dt} = V$

given that.

I = 0.3A

R= 30 $\ohm$ohm

so V = 0.3 * 30 = 9 volts

$V = {500}\frac{d(AB) }{dt} = 9 volts$

$V = {0.00196 \times 500}\frac{d(B) }{dt} = 9 volts$

${0.845}\frac{d(B) }{dt} = 9 volts$

$\frac{d(B) }{dt} = 10.65$