VITEEE 2015 :: Physics :: General questions

• Physics - General questions

The value of inductance L for which the current is maximum  in series LCR circuit with C=10 $\mu$ F  and ω= 1000 rad/s

Answer:

Explanation:

 Maximum current flows in the circuit in resonance condition Current in the LCR circuit 

$i=\frac{V}{\sqrt{R^{2}+(X_{L}-X_{C})^{2}}}$

 For current to be maximum denominator should be minimum

$(X_{L}-X_{C})^{2}$= 0

$\Rightarrow X_{L}=X_{C}\Rightarrow\omega L=\frac{1}{\omega C}$

$\therefore$   $ L=\frac{1}{\omega^{2}C}=\frac{1}{(100)^{2}\times10\times 10^{-6}}$

$ L=\frac{1}{10}H=0.1 H=100 mH$

 A coil of area 5 cm² having 20 turns is placed in a uniform magnetic field of 10³ gausses. The normal the plane of coil makes an angle 30° with the magnetic field. The flux through the coil is 

Answer:

Explanation:

Given :N=20

 B= 10³ gauss= 10³ x 10^-4 T= 0.1T

 A= 5 cm2= 5 x 10^-4 m²

 $\theta$= 80°

 Flux through the coil =   

$\phi= NBA \cos\theta$

$=20\times 0.1\times 5\times 10^{-4}\times\cos 30^{0}$

 $=10\times 10^{-4}\times \frac{\sqrt{3}}{2}=5\sqrt{3}\times 10^{-4}=865\times 10^{-4}wb$

An induced emf has

Answer:

Explanation:

 From Lentz's law, the direction of induced emf in a circuit is such that it opposes the magnetic flux that produces it.

  So, if the magnetic flux linked with closed-circuit increases the induced current flows in a direction so as to develop a magnetic flux in the opposite direction of the original flux.

 If the magnetic flux linked with a closed circuit decreases then the induced current flows in the same direction of the original flux. So, the induced emf has not the direction of its own 

A magnetic needle lying parallel to the magnetic field requires W units of work to turn it through an angle of 45°. The torque required to maintain the needle in this position will be

Answer:

Explanation:

 Work done by magnet to turn from angle  $\theta_{1}$ to $\theta_{2}$

 $W=MB(\cos\theta_{1}-\cos\theta_{2})$

$=MB(\cos0^{0}-\cos 45^{0})$

$W=MB(1-\frac{1}{\sqrt{2}})=(\frac{\sqrt{2}-1}{\sqrt{2}})MB$

 Also torque acting on the magnet

 $\tau=MB\sin 45^{0}=\frac{MB}{\sqrt{2}}$

$W=(\sqrt{2}-1).\tau\Rightarrow\tau=\frac{W}{(\sqrt{2}-1)}$

Which of the following statement related to the hysteresis loop is incorrect?

Answer:

Explanation:

 The hysteresis loop, i.e,  area of B-H curve is a measure of energy dissipated per cycle per unit volume of the specimen. It depends on the nature of magnetic material

• Physics - General questions