CBSE Questions for Class 12 Medical Physics Electromagnetic Induction Quiz 8 - MCQExams.com

The correct answer is option (B).

Hint: Calculate the magnetic flux and use it to find out the mutual inductance.

Step 1: Calculate magnetic flux of smaller loop.

Let the length of the larger square and the smaller square be $$L$$ and $$l$$ respectively. Accordingly, the current in these two square loops are $$I$$ and $$i$$.

According to Faraday’s first law of electromagnetic radiation, whenever magnetic flux passes through surface changes with respect to time, an induced EMF is generated.
Magnetic field at a distance $$\dfrac{L}{2}$$ from the centre of the current carrying wire of length $$L$$ is given by:

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i}{\dfrac{L}{2}}2sin 45$$

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$$

Magnetic field in the centre of square due to four wires of length $$L$$ will be:

$$B_1=\dfrac{\mu _0}{4\pi}\dfrac{4\times i\times 2\times 2sin \times \dfrac{1}{\sqrt{2}}}{L}$$

$$B_1=\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2i}{L}$$

Since, the smaller loop is at the centre of the larger loop, so $$L>>l$$

In a small square, the magnetic field is uniform and is equal to $$B_1$$

Flux in the smaller loop will be:

$$\phi =B_1l^2$$

$$\phi =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2il^2}{L}$$

Step 2: Calculate the mutual inductance.

Mutual inductance is given by –

$$M=\dfrac{\phi}{i}$$

$$M =\dfrac{\mu_0}{\pi}\dfrac{2\sqrt2l^2}{L}$$

$$M\propto \dfrac{l^2}{L}$$

Hence, the mutual inductance of the system is proportional to $$\dfrac{l^2}{L}$$.

The correct answer is option (B).

Given,

Angular frequency $$\omega =2\pi f=\dfrac{2\pi 1800}{60}=60\pi $$

Magnetic field, $$B=0.5\ G=0.5\times {{10}^{-4}}\ T$$

Area, $$A=\pi \times {{0.07}^{2}}=0.0154\,{{m}^{2}}$$ 

Flux $$\phi =nBA\cos \theta =nB(\pi {{r}^{2}})\cos \omega t$$

The induced emf is

$$\varepsilon =\dfrac{-d\phi }{dt}=nBA\omega \sin \omega t$$

For maximum E.M.F, $$\varepsilon ={{\varepsilon }_{o}},\ \sin \omega t=1$$

  $$ {{\varepsilon }_{o}}=n\omega BA $$

 $$ {{\varepsilon }_{o}}=4000\times 60\pi \times 0.5\times {{10}^{-4}}\times 0.0154 $$

 $$ {{\varepsilon }_{o}}=0.58\,V $$ 

Given,

Velocity,  $$V=10\,cm/s$$

Length of rod, $$L=50\,cm$$

Magnetic Field, $$B=10\,G$$

Motional $$emf=BlV\sin \theta =10\times 50\times 10\sin {{30}^{o}}=2500\,\ CGS\ unit$$ 

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 $$4\ J$$

Given,

The disc is rotating in magnetic field from North to South.

If the direction of the magnetic field is perpendicular to the plane of the disc:

Viewing disk from the top.

Force on each proton is toward the inside, and force on each electron is outside.

So, the negative charge will develop, in the outskirt of the disk, and the positive charge will develop at the center.

Hence, Current will flow from P to Q. 

$$Alternatively:$$

If the magnetic field is perpendicular to the normal vector of the plane of the disc:

The emf induced will not be radial, as governed by flaming's right hand rule, thus no current will flow through the resistance R.

The rate of change of magnetic flus is called as induced emf. To induce an emf, magnetic field must be changing. When the flux in the  graph is constant then there is no magnetic field and hence no emf induces.

Induced emf is negative times the rate of change of flux. 

So, induce emf is shown in graph 3

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: 

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