Tag: electromagnetic induction

Questions Related to electromagnetic induction

The sun delivers ${10^3}W/{m^2}$ of electromagnetic flux to the earth's surface. The solar energy incident on the roof in $1$houre will be 

  1. $5.76 \times {10^8}J$

  2. $5.76 \times {10^7}J$

  3. $5.76 \times {10^6}J$

  4. $5.76 \times {10^5}J$


Correct Option: C

Light with an energy flux of $18 w/cm^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 cm^2$. Find the average force exerted on the surface during a 30 minute time

  1. $1.2 \times 10 ^ { - 6 } N$

  2. $2 .4\times 10 ^ { - 6 } N$

  3. $2.16 \times 10 ^ { - 3 } \mathrm { N }$

  4. $1.5\times 10 ^ { - 6 } N$


Correct Option: A
Explanation:

The total energy falling on the surface is $U = \left( {18W/c{m^2}} \right) \times \left( {20c{m^2}} \right) \times \left( {30 \times 60} \right) = 6.48 \times {10^5}J$

therefore$,$ the total momentum delivered is 
$P = \dfrac{U}{c} = \dfrac{{\left( {6.48 \times {{10}^5}J} \right)}}{{\left( {3 \times {{10}^8}\,m/s} \right)}} = 2.16 \times {10^{ - 3}}\,kgm/s$
The average force exerted on the surface is 
$F = \dfrac{p}{t} =  = \dfrac{{2.16 \times {{10}^{ - 3}}}}{{0.18 \times {{10}^4}}} = 1.2 \times {10^{ - 6}}N$
Hence,
option $(A)$ is correct answer.

The electric field in a certain region is $\left( 10\hat { i } +5\hat { j }  \right) \times { 10 }^{ 4 }N/C$. What is the flux due to this field over an area of $\left( 3\hat { i } +3\hat { j }  \right) \times { 10 }^{ -2 }{ m }^{ 2 }$ in ${ Nm }^{ 2 }/C?$

  1. $4.5\times { 10 }^{ 3 }$

  2. $3.5\times { 10 }^{ 3 }$

  3. $2.5\times { 10 }^{ 3 }$

  4. $1.5\times { 10 }^{ 3 }$


Correct Option: A

Current flowing through a long solenoid is varied. Then, magnetic flux density of the magnetic field inside varies ::

  1. inversely with $I$

  2. inversely with ${ I }^{ 2 }$

  3. directly with $I$

  4. directly with ${ I }^{ 2 }$


Correct Option: C

A wire of length $'\ell '$ is used to make a circular coil of 'n' no. of turns. A current 'I' passes through the coil. If twice the length of wire is used now to make the coil with turns of same radius, making same current flow through it, the magnetic field at the centre of coil will become.

  1. 1.5 times than earlier

  2. 0.5 time than earlier

  3. 2 times than earlier

  4. 4 time than earlier


Correct Option: B

The magnetic flux linked with a coil is given by the equation $ \phi = 5t^2 + 3t + 6 $. The induced e.m.f. in the coil in the fourth second will be 

  1. 20 V

  2. 40 V

  3. 10 V

  4. 80 V


Correct Option: C

A closely wound flat circular coil of 25 turns of wire has diameter of =f 10 cm and carries a current of 4 amperes. determine the magnetic flux density at the centre of the coil:-

  1. $ 1.679 \times 10^{-5} T $

  2. $ 2.028 \times 10^{-4} T $

  3. $ 1.257 \times 10^{-3} T $

  4. $ 1.512 \times 10^{-6} T $


Correct Option: C

The magnetic flux linked with a coil, in webers, is given by the equation $\phi =4t^{2}-3t+7$. Then the magnitude of induced emf at 2 sec will be....

  1. 15 v

  2. 19 v

  3. 17 v

  4. 21 v


Correct Option: C

Magnetic field intensity at the centre of coil of 50 turns, radius 0.5 m and carrying a current of 2 A is 

  1. $0.5 \times 10^{-5}T$

  2. $3 \times 10^{-5}T$

  3. $1.25 \times 10^{-4}T$

  4. $4 \times 10^{-5}T$


Correct Option: C

Select the incorrect option

  1. Luminous flux and radiant flux have same dimensions

  2. Luminous flux and luminous intensity have same dimensions

  3. Radiant flux and power have same dimension

  4. Relative luminosity is a dimensionless quantity


Correct Option: A
Explanation:

Luminous flux ( Luminous power) is the amount of perceived power of light. It is measure energy.

Radiant flux (Radiant power) is the amount of radiated power.
While luminous intensity is ratio of luminous flux to solid angle.