Electromagnetic Induction Test - 54

Given that,

Self-inductance of coil = $$2.0 H$$

Let the current flowing through the coil be 2A at time $${{t}^{2}}$$

Now,

  $$ 2=2\sin {{t}^{2}} $$

 $$ t=\sqrt{\dfrac{\pi }{2}} $$

Now, the amount of energy spent during the period when the current changes from $$0\ to\ 2\ A$$

Now, Self-induced e. m. f

$$E=L\dfrac{di}{dt}$$

Now, the work done is

  $$ dW=L\left( \dfrac{di}{dt} \right)dq $$

 $$ dW=L\times \dfrac{di}{dt}\times idt $$

 $$ dW=Lidi $$

Now, on integrate

  $$ \int{dW=\int\limits_{0}^{t}{Li}}di $$

 $$ W=\int\limits_{0}^{t}{L2\sin {{t}^{2}}d\left( 2\sin {{t}^{2}} \right)} $$

 $$ W=\int\limits_{0}^{t}{8L\sin {{t}^{2}}\cos {{t}^{2}}tdt} $$

 $$ W=4L\int\limits_{0}^{t}{\sin 2{{t}^{2}}dt} $$

Now, let

  $$ \theta =2{{t}^{2}} $$

 $$ d\theta =4tdt $$

 $$ dt=\dfrac{d\theta }{4t} $$

Now, put the value of $$2{{t}^{2}}$$and $$dt$$

  $$ W=4L\int\limits_{0}^{t}{\dfrac{\sin \theta d\theta }{4}} $$

 $$ W=L\left[ \left( -\cos \theta  \right) \right]_{0}^{t} $$

 $$ W=-L\left[ \cos 2{{t}^{2}} \right]_{0}^{t} $$

 $$ W=-L\left[ \cos 2{{t}^{2}} \right]_{0}^{\sqrt{\dfrac{\pi }{2}}} $$

 $$ W=2L $$

 $$ W=2\times 2 $$

 $$ W=4\,J $$

Hence, the amount of energy is $$4\ J$$

When both coil and magnet moves in same direction, then there will be no induced emf in the coil because there is no change in the magnetic flux. If both are moves in same direction, then induced emf in coil is zero. The induced emf is given by: 

$$emf=\dfrac{-d\phi }{dt}=\dfrac{-d(BA)}{dt}$$

Given that,

Radius $$r=2\,cm$$  

$$\dfrac{dB}{dt}=2\,T$$

We know that

Magnetic flux

$$ \phi =BA $$

$$ \phi =B\left( \pi {{r}^{2}} \right) $$

Now, induced e. m.f

 $$ E=\dfrac{d\phi }{dt} $$

 $$ E=\dfrac{d\left( B\pi {{r}^{2}} \right)}{dt} $$

 $$ E=\pi {{r}^{2}}\dfrac{dB}{dt} $$

 $$ E=3.14\times {{\left( 0.02 \right)}^{2}}\times 2 $$

 $$ E=0.002\,V $$

Hence, the induced e. m. f is $$0.002\ V$$