Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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. $1.5\times 10 ^ { - 6 } N$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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. $1.5\times { 10 }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electric flux is the dot product of the electric field vector and the area vector. Phi = (10i + 5j) * 10^4 * (3i + 3j) * 10^-2 = (30 + 15) * 10^2 = 45 * 10^2 = 4.5 * 10^3 Nm^2/C.

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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 ${ I }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The magnetic field inside a long solenoid is given by B = mu0 * n * I, where n is the number of turns per unit length. Thus, B is directly proportional to the current I.

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a flat circular coil, B at center = μ₀NI/(2r). Given: N = 25 turns, I = 4 A, r = 5 cm = 0.05 m. B = (4π × 10⁻⁷ × 25 × 4)/(2 × 0.05) = 4π × 10⁻⁷ × 100/0.1 = 4π × 10⁻⁴ ≈ 1.257 × 10⁻³ T. This matches option C. The formula differs from solenoid (which uses turns per length).

Multiple choice physics magnetic fields and electromagnetism magnetic flux density magnetic flux electromagnetic induction

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. $4 \times 10^{-5}T$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

B = (mu0 * N * I) / (2 * R) = (4 * pi * 10^-7 * 50 * 2) / (2 * 0.5) = (4 * pi * 10^-7 * 100) / 1 = 4 * pi * 10^-5 = 1.256 * 10^-4 T.