Tag: electromagnetic forces

Quizzes (4) Questions (60)

Questions Related to electromagnetic forces

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

When the energy of the incident radiation is increased by 20%, The kinetic energy of the photelectrons emitted from a metal surfce in creased from 0.5V to 0.8eV The work function of the metal is 

  1. 0.65 eV

  2. 1.0 eV

  3. 1.2 eV

  4. 1.5 eV
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Einstein's photoelectric equation, K = E - W. Let initial energy be E and final energy be 1.2E. We have 0.5 = E - W and 0.8 = 1.2E - W. Solving these simultaneous linear equations gives W = 1.0 eV.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

Out of the following transitions, the frequency of emitted photon will be maximum for 

  1. $n = 5$ to $n = 3$
  2. $n = 6$ to $n = 2$
  3. $n = 2$ to $n = 1$
  4. $n = 1$ to $n = 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The frequency of an emitted photon is proportional to the energy difference of the transition (E = hf). The energy difference is greatest for the n=2 to n=1 transition in the hydrogen atom, as the energy levels are spaced further apart at lower quantum numbers.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

When electron is incident on molybdenum they by changing energy of electron:-

  1. $\lambda _{\text{min}}$ changes
  2. $\lambda _{\text{min}}$ remains constant
  3. $\lambda _{K _{\alpha}}, \lambda _{K _{\beta}}$ changes
  4. $\lambda _{\text{min}}, \lambda _{K _{\alpha}}$ and $\lambda _{K _{\beta}}$ all changes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The minimum wavelength (lambda_min) of X-rays produced is determined by the accelerating voltage (or energy) of the incident electrons via the Duane-Hunt law, lambda_min = hc/eV. Characteristic X-ray wavelengths (K_alpha, K_beta) depend on the target material (molybdenum), not the incident electron energy.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

An electron beam moving with a speed of $2.5\times 10^{7}ms^{-1}$  enters into the magnetic field directed perpendicular to its direction of motion. The magnetic induction of the field is $4\times 10^{-3}  wb/m^2$. The intensity of the electric field applied so that the electron remains undeflected due to the magnetic field is.

  1. $10^{4}N/C$
  2. $10^{5}N/C$
  3. $10^{7}N/C$
  4. $10^{3}N/C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no deflection
VB = E
$\rightarrow E = 2.5*10^7*4*10^{-3}$
$= 10*10^4$
$= 10^5 {N}/{C}$

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

J.J. Thomson's cathode-ray tube experiment demonstrated that:

  1. cathode rays are streams of negatively charged ions

  2. all the mass of an atom is essentially in the nucleus

  3. the e/m of electrons is much greater than the e/m of protons

  4. the e/m ratio of the cathode ray particles changes when a different gas is placed in the discharge tube
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

(A) is the correct option. Since, in Thomson's first experiment, he discovered the cathode rays and the change they deposited were intrinsically linked together. In the second experiments, he discovered that the cathode ray were negative. He deducted that the cathode rays were made up of negatively charged particles.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

A particle of mass $1\times 10^{-26} \ kg $ and charge $1.6\times 10^{-19} \ C$ travelling with a velocity $1.28\times 10^6 \ m/s$ along the positive $X$-axis  enters a region in which a uniform electric field $\vec E$ and a uniform magnetic field of induction $\vec B$ are present. if $\vec E=-102.4 \times 10^3 \ \hat k \ NC^{-1}$ and $\vec B=8 \times 10^{-2} \ \hat j \ Wbm^{-2}$, the direction of motion of the particles is 

  1. along the positive X-axis

  2. along the negative X-axis
  3. at $45^{O}$ to the positive X-axis
  4. at $135^{O}$to the postive X-axis
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Write Force
F = q(vXB) + qE.
F =0N.
Hence the particle will continue moving towards positive X-axis.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

A proton and an $\alpha$ particle enter a magnetic field in a direction perpendicular to it. If the force acting on the proton is twice that acting on the $\alpha$- particle, the ratio of their velocities is

  1. $4 : 1$
  2. $1 : 4$
  3. $1 : 2$
  4. $2 : 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that Force $(F _m)$ on a particle with charge "q" moving with a velocity of v at an angle $\theta$ to the magnetic field $(B)$ can be expressed as

$F _m=qvB \sin{\theta}$
Given, $\cfrac{F _m (p)}{F _m(\alpha)}$=2
$q _{\alpha}=2 \times q _{p}$
$\Rightarrow \ \cfrac{q _{p}v _pB}{q _{\alpha}v _{\alpha}B}=2$
We get 
$\cfrac{v _{p}}{v _{\alpha}}=4$

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

When cathode rays moving horizontally pass through an electric field directed vertically downward, the rays would be deflected ________ .

  1. downward

  2. upward

  3. backward

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is directed vertically downward i.e. the upper plate is positively charged and the lower plate is negatively charged . Because $ \vec{E} $ points in a direction from positive to negative.

When a cathode ray particle  i.e. negatively charged particle enters the electric field, it will be following a parabolic trajectory and would be directed towards the  positive plate. Since the positive plate is upward directed. So cathode ray will be directed upwards.

Multiple choice deflection of electron beam by magnetic field observing the force and electron beam tubes charged particles electromagnetic forces physics

When a charged particle of charge $x\ C$ moves through a potential difference of $y\ V$, the gain in kinetic energy is equal to $xy \ J$.
An electron and an alpha particle have their masses in the ratio of 1 : 7200 and charges in the ratio of 1 : 2. If they start moving from rest through the same electrical potential difference then the ratio of their velocities will be ______. 

  1. 1 : 60

  2. 60 : 1

  3. 1 : 20

  4. 20 : 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given : If particle has charge $x$  coulombs and potential  difference $ y $ volts . Then gain in K.E. = $ xy $ joules.

Using similar analysis,

$ \dfrac{\text{(mass)electron}}{\text{(mass) alpha particle}} = \dfrac{1}{7200} $

$ \dfrac{\text{(charge)electron}}{\text{(charge) alpha particle}} = \dfrac{1}{2} $

We need to find $ \dfrac{\text{(velocity)electron}}{\text{(velocity) \alpha particle}} $

Since initially there at rest, initial K.E. = 0;

They move through same potential difference. Let potential difference be $ y $ volts.

So, gain in K.E = Final K.E – Initial K.E

= Final K.E – 0

So, gain in K.E = $ charge \times potential  \, difference $

Ratio of gain in K.E of electron and $\alpha$ particle = Ratio of final K.E of electron and $\alpha$ particle

 

$ \dfrac{\dfrac{1}{2} m v _e^2 \times y}{\dfrac{1}{2} m v _\alpha^2 \times y} $

 

$ = \dfrac{1}{2} \times 7200 = 3600 $

$ \dfrac{v _e}{v _\alpha} = \sqrt{3600} = 60 $

 

 

 

Multiple choice physics current electricity and magnetism electric bell the magnetic effect of a current electromagnetic forces oersted experiment oersted's experiment magnetic field due to a straight current carrying conductor

What material is used for making the armature of an electric bell ? Give a reason for using this material.

  1. Soft Iron

  2. Carbon

  3. Glass

  4. Silicon
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Soft iron is used for making the armature of an electric bell because soft iron is magnetised easily on passing the current and completely demagnetised easily on withdrawing the current through the solenoid.