VITEEE 2015 :: Physics :: General questions

• Physics - General questions

The earth is considered as a short magnet with its centre coinciding with the geometric centre of earth. The angle of dip $\phi$ related to the magnetic latitude  $\alpha\lambda$  as

Answer:

Explanation:

 For a dipole at position (R, Q)

 $B_{R}=\frac{\mu_{0}}{4\pi}.\frac{2M\cos\theta}{R^{3}}$   .....(i)

 and    $B_{Q}=\frac{\mu_{0}}{4\pi}.\frac{M\sin\theta}{R^{3}}$

 also   $\tan\phi=\frac{B_{V}}{B_{H}}=-\frac{B_{R}}{B_{Q}}$   ......(iii)

 Dividing eq.(i) by (ii)

 $\frac{B_{R}}{B_{Q}}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta$    .....(iv)

 From eq.(iii) and (iv)

 $\tan\phi=-2\cot\theta$

From figure ,  $\theta=90^{0}+\lambda$

 $\because$     $\tan\phi=-2\cot(90+\lambda)$

$\tan\phi=2\tan\lambda$

A galvanometer with a scale  divided into 100 equal  divisions has a current sensitvity  of 10 divisions per milliampere   and a voltage  sensitivity  of 2 divisions per millivolt. The galvanometer resistance will be 

Answer:

Explanation:

 Given : current  sensitvity =10 div/mA and there are 100 division on the scale .

 $\therefore$ Current required for full scale deflection

 $I_{g}=\frac{1}{10}\times 100 mA=10 mA=0.01 A$

Also voltage sensitivity =2 div/mV

$\therefore$   Voltage required for  full deflection 

  $V{g}=\frac{1}{2}\times 100 mV=0.05 V$

 Galvanometer resistance is given by

 $G=\frac{V_{g}}{I_{g}}=\frac{0.05}{0.01}=5$Ω

The direction of magnetic field dB due to current element dl at a distance r is the  direction of 

Answer:

Explanation:

 The direction of dB is the direction of vector dl x r From right-hand screw rule if we place a right-handed screw at the point where the magnetic field is needed to be determined and turn its handle from dl to r. then the direction in which the  screw advances gives the direction of field dB

A current of 2 A is flowing in the sides of an equilateral triangle of side 9 cm. The magnetic field at the centroid  of the triangle is 

Answer:

Explanation:

 Due  to current through side AB Magnetic field at the centre O

$B_{1}=\frac{\mu_{0}I}{4\pi a}[\sin \theta_{1}+\sin \theta_{2}]$

 As the  magnetic field due to each of the three sides is the same in magnitude  and direction

 $\therefore$ Total magnetic field at O  is the sum of all the fields.

 i.e,   $B=3B_{1}=\frac{3\mu_{0}I}{4\pi a}[\sin \theta_{1}+\sin \theta_{2}]$

 Here,    $\tan\theta_{1}=\frac{AD}{OD}\Rightarrow\tan 60^{0}=\frac{\frac{l}{2}}{a}$

$a=\frac{l}{2\sqrt{3}}=\frac{9\times 10^{-2}}{2\sqrt{3}}$

 Now B

$=3\times\frac{4\pi\times 10^{-7}\times2}{4\pi\times\frac{9\times 10^{-2}}{2\sqrt{3}}}[\sin 60^{0}+\sin 60^{0}]$

 =  $\frac{4\sqrt{3}}{9}\times 10^{-5}[\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}]$

 =1.33 x 10^-5 T

 A horizontal rod of mass 0.01 kg and length 10 cm is placed on a friction less plane inclined at an angle 60° with trhe horizontal and with the  length  of rod parallel to the edge of the inclined  plane. A uniform magnetic  fiewld is applied 'Vertically downwards. If the current through the rod  is 1.73 A, then the value of magnetic field induction B  for which  the rod remains stationary  on the inclined  plane is 

Answer:

Explanation:

 Here two forces acting on the rod simultaneously

 From FBD, mg sing 60 = Bil cos 60°

 $B=\frac{mg}{il}\tan 60^{0}=\frac{0.01\times10}{173\times0.1}\times\sqrt{3}=1T$

• Physics - General questions