An electron enters a magnetic field at right angle to it as shown in the figure. The direction of the force acting on the electron will be:-

We know that both the directions are perpendicular, thus for force direction=?Using Fleming's left hand rule,Direction of force is perpendicular to the direction of magnetic field and current. Thus direction of force is opposite to electron motion into the page at \\(90°\\)

A proton, a deuteron and an \\(\alpha-particle\\) with the same kinetic energy enter a region of uniform magnetic field moving at right angles to \\(B\\). What is the ratio of the radii of their circular paths?

\\(\dfrac {mv^{2}}{R} = qvB\\)For proton, \\(R_{p} = \dfrac {mv}{Bq} = \dfrac {\sqrt {2M_{p}E}}{q_{p}B}\\)Similarly for deuteron and \\(\alpha-particle\\)\\(R_{d} = \dfrac {\sqrt {2M_{d}E}}{q_{d}B}\\) and \\(R_{\alpha} = \dfrac {\sqrt {2M_{\alpha}E}}{q_{\alpha}B}\\)According to the question\\(\therefore R_{p} : R_{d} : R_{\alpha}\\)or \\(\dfrac {\sqrt {M_{p}}}{q_{p}} : \dfrac {\sqrt {M_{d}}}{q_{d}} : \dfrac {\sqrt {M_{\alpha}}}{q_{\alpha}}\\)\\(\therefore \dfrac {\sqrt {1}}{1} : \dfrac {\sqrt {2}}{1} : \dfrac {\sqrt {4}}{2}\\)or \\( {1} : \sqrt {2} : 1\\).

A proton, a deuteron and an \\(\alpha\\)-particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths described by them is

Radius of circular path \\(r = \dfrac{mv}{Bq}\\)Using \\(p=mv = \sqrt{2mK}\\)where \\(K\\) is the kinetic energy of the particle.We get \\(r = \dfrac{\sqrt{2mK}}{Bq}\\)\\(\implies\\) \\(r \propto \dfrac{\sqrt{m}}{q}\\)Given : \\(q_{\alpha} = 2q_p\\) and \\(q_d = 2q_p\\) where \\(q_p\\) is the charge of proton.Also, \\(m_{\alpha} =4m_p\\) and \\(m_d = 2m_p\\) where \\(m_p\\) is the mass of proton.\\(\therefore\\) \\(r_p : r_d:r_{\alpha} = \dfrac{\sqrt{m_p}}{q_p} :\dfrac{\sqrt{m_d}}{q_d}: \dfrac{\sqrt{m_{\alpha}}}{q_{\alpha}}\\)Or \\(r_p : r_d:r_{\alpha} = \dfrac{\sqrt{m_p}}{q_p} :\dfrac{\sqrt{2m_p}}{2q_p}: \dfrac{\sqrt{4m_p}}{2q_{p}}\\)\\(\implies\\) \\(r_p : r_d:r_{\alpha} =\sqrt2:1:\sqrt{2}\\)

Two particles X and Y having equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe a circular paths of radii '\\(r_1\\) and '\\(r_2\\)' respectively. The ratio of the mass of X to that of Y is

Radius of circular path traced by a particle \\(r = \dfrac{mv}{Bq}\\)Kinetic energy \\(K = \dfrac{1}{2}mv^2 = eV\\)\\(\therefore\\) \\(r = \dfrac{\sqrt{m^2 v^2}}{Bq} = \dfrac{\sqrt{2mK}}{Bq}\\)Or \\(r = \dfrac{\sqrt{2meV}}{Bq}\\)\\(\implies \\) \\(m \propto r^2\\)Thus we get the ratio of masses of \\(X\\) to \\(Y\\) \\(\dfrac{m_1}{m_2} = \bigg[\dfrac{r_1}{r_2} \bigg]^2\\)

A proton and an \\(\alpha\\)-particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths described by the proton and \\(\alpha\\)-particle is

When a charged particle enters a field, it takes a circular path. The radius of circular path is given by the relation\\(\displaystyle r=\frac{mv}{Bq}\\)For proton, \\(r_1=\displaystyle\frac{mv}{Bq}\\)For \\(\alpha\\)-particle, \\(r_2=\displaystyle\frac{4m\times v}{B\times 2q}=\frac{2mv}{Bq}\\)\\(\Rightarrow r_1: r_2 : : 1:2\\)

A particle of mass m, charge \\(q\\) and kinetic energy \\(t\\) enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

after passing through a magnetic field, the magnitude of its mass and velocity of the particle remains same, so its energy does not change that is kinetic energy will remain T.Hence (C) is correct answer

A uniform magnetic field exists in the plane of paper pointing from left to right as shown in figure. In the field, an electron and a proton move as shown. The electron and the proton experiences:

For the given figure, the proton and electron are moving in opposite directions to each other and is perpendicular to the direction of magnetic field. Now, we know that the direction of current is taken opposite to the direction of motion of electron.So, both electron and proton have current in same direction. Therefore, the forces acting on them are given by Fleming s left hand rule and they are pointing into the plane of the paper.

A proton, a deuteron and an \\(\alpha-particle\\) having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If \\(r_p, r_d\\) and \\(r_{\alpha}\\) denote, respectively, the radii of the trajectories of these particles, then

The centripetal force is provided by the magnetic force\\(\dfrac {mv^2}{r}=qvB\Rightarrow \dfrac {mv}{r}=qB\\)\\(\Rightarrow r=\dfrac {mv}{qB}=\dfrac {p}{qB}[\because p-mv]\\)But \\(KE=\dfrac {p^2}{2m}\Rightarrow p=\sqrt {2mKE}\\)Here, KE and B are same for the three particles\\(\therefore r\propto \dfrac {\sqrt m}{q}\\)\\(\therefore r_p : r_d : r_{\alpha}=\dfrac {\sqrt 1}{2} : \dfrac {\sqrt 2}{1} :\dfrac {\sqrt 4}{2}=1:\sqrt 2:1\\)\\(\Rightarrow r_{\alpha}=r_p < r_d\\).

An electron is moving with a speed of \\(3 \times 10^{-7} m/s\\) in a magnetic field of \\(6 \times 10^{-4} T\\) perpendicular to its path. What will be the radius of the path ? What will be frequency and the energy in keV ?[Given: mass of electron \\(= 9 \times 10^{-31}kg, charge \,e = 1.6 \times 10^{-19}C, eV = 1.6 \times 10^{-19} \,J\\)]

Here, \\(m = 9 \times 10^{-31}kg,\\)\\(q = 1.6 \times 10^{-19}C, v = 3 \times 10^7 \,ms^{-1},\\)\\(b = 6 \times 10^{-4} T\\)\\(r = \dfrac{mv}{qB} = \dfrac{ (9 \times 10^{-31}) \times (3 \times 10^7)}{(1.6 \times 10^{-19}) (6 \times 10^{-4} )} = 0.28 \,m\\)\\(v = \dfrac{v}{2\pi r} = \dfrac{Bq}{2\pi m} = \dfrac{ (6 \times 10^{-4}) \times (1.6 \times 10^{-19})}{2 \times (22/7) \times 9 \times 10^{-31}}\\)\\(= 1.7 \times 10^7 \,Hz\\)\\(Ek = \dfrac{1}{2} mv^2 =\\)\\(\dfrac{1}{2} \times (9 \times 10^{-31}) \times (3 \times 10^7)^2 J\\)\\(= 40.5 \times 10^{-17}J = \dfrac{40.5 \times 10^{-17}}{1.6 \times 10^{-16}} keV\\)\\(= 2.53 \,keV\\)

Motion in the Presence of Magnetic Field | Definition, Examples, Diagrams

Q: An electron is moving with a speed of \\(3 \times 10^{-7} m/s\\) in a magnetic field of \\(6 \times 10^{-4} T\\) perpendicular to its path. What will be the radius of the path ? What will be frequency and the energy in keV ?[Given: mass of electron \\(= 9 \times 10^{-31}kg, charge \,e = 1.6 \times 10^{-19}C, eV = 1.6 \times 10^{-19} \,J\\)]

Here, \\(m = 9 \times 10^{-31}kg,\\)\\(q = 1.6 \times 10^{-19}C, v = 3 \times 10^7 \,ms^{-1},\\)\\(b = 6 \times 10^{-4} T\\)\\(r = \dfrac{mv}{qB} = \dfrac{ (9 \times 10^{-31}) \times (3 \times 10^7)}{(1.6 \times 10^{-19}) (6 \times 10^{-4} )} = 0.28 \,m\\)\\(v = \dfrac{v}{2\pi r} = \dfrac{Bq}{2\pi m} = \dfrac{ (6 \times 10^{-4}) \times (1.6 \times 10^{-19})}{2 \times (22/7) \times 9 \times 10^{-31}}\\)\\(= 1.7 \times 10^7 \,Hz\\)\\(Ek = \dfrac{1}{2} mv^2 =\\)\\(\dfrac{1}{2} \times (9 \times 10^{-31}) \times (3 \times 10^7)^2 J\\)\\(= 40.5 \times 10^{-17}J = \dfrac{40.5 \times 10^{-17}}{1.6 \times 10^{-16}} keV\\)\\(= 2.53 \,keV\\)

definition

Aurora Boreails

Interaction of the terrestrial magnetic field with particles from the solar wind sets up the conditions for the aurora phenomena near the poles.

The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earths atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.

Auroras that occur in the northern hemisphere are called Aurora Borealis or northern lights and auroras that occur in the southern hempishere are called Aurora Australis or southern lights.

definition

Motion of a particle in a magnetic field

The particle will describe a circle if

v

and

B

are perpendicular to each other .If velocity has a component along

B

, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to

B

is as before a circular one, thereby producing a helical motion

example

Energy of a charged particle moving in a magnetic field

Calculate the energy in keV of an electron (mass

9 \times 1 0^{- 31} k g

and charge

1.6 \times 1 0^{- 19} C

) moving at a speed of

3 \times 1 0^{7} m / s

in a magnetic field of

6 \times 1 0^{- 4} T

perpendicular to it

E = \frac{1}{2} m v^{2} = \frac{1}{2} 9 \times 1 0^{- 31} k g \times 9 \times 1 0^{14} m^{2} / s^{2} = 2.5 k e V

Motion in the Presence of Magnetic Field

definition

Aurora Boreails

definition

Motion of a particle in a magnetic field

example

Energy of a charged particle moving in a magnetic field

REVISE WITH CONCEPTS

Motion in the Presence of Magnetic Field - Problems L1

Quick Summary With Stories

Charged Particle in Uniform Magnetic Field - 1

Charged Particle in Uniform Magnetic Field - 2

An electron enters a magnetic field at right angle to it as shown in the figure. The direction of the force acting on the electron will be:-

A proton, a deuteron and an $α - p a r t i c l e$ with the same kinetic energy enter a region of uniform magnetic field moving at right angles to $B$ . What is the ratio of the radii of their circular paths?

A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is in which direction?

A proton, a deuteron and an $α$ -particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths described by them is

Two particles X and Y having equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe a circular paths of radii ' $r_{1}$ and ' $r_{2}$ ' respectively. The ratio of the mass of X to that of Y is

A proton and an $α$ -particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths described by the proton and $α$ -particle is

A particle of mass m, charge $q$ and kinetic energy $t$ enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

A uniform magnetic field exists in the plane of paper pointing from left to right as shown in figure. In the field, an electron and a proton move as shown. The electron and the proton experiences:

A proton, a deuteron and an $α - p a r t i c l e$ having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_{p}, r_{d}$ and $r_{α}$ denote, respectively, the radii of the trajectories of these particles, then

definition

Aurora Boreails

definition

Motion of a particle in a magnetic field

example

Energy of a charged particle moving in a magnetic field

REVISE WITH CONCEPTS

Motion in the Presence of Magnetic Field - Problems L1

Quick Summary With Stories

Charged Particle in Uniform Magnetic Field - 1

Charged Particle in Uniform Magnetic Field - 2

An electron enters a magnetic field at right angle to it as shown in the figure. The direction of the force acting on the electron will be:-

A proton, a deuteron and an α−particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to B. What is the ratio of the radii of their circular paths?

A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is in which direction?

A proton, a deuteron and an α-particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths described by them is

Two particles X and Y having equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe a circular paths of radii 'r1​ and 'r2​' respectively. The ratio of the mass of X to that of Y is

A proton and an α-particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths described by the proton and α-particle is

A particle of mass m, charge q and kinetic energy t enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

A uniform magnetic field exists in the plane of paper pointing from left to right as shown in figure. In the field, an electron and a proton move as shown. The electron and the proton experiences:

A proton, a deuteron and an α−particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp​,rd​ and rα​ denote, respectively, the radii of the trajectories of these particles, then

An electron is moving with a speed of 3×10−7m/s in a magnetic field of 6×10−4T perpendicular to its path. What will be the radius of the path ? What will be frequency and the energy in keV ? [Given: mass of electron =9×10−31kg,chargee=1.6×10−19C,eV=1.6×10−19J]

A proton, a deuteron and an $α - p a r t i c l e$ with the same kinetic energy enter a region of uniform magnetic field moving at right angles to $B$ . What is the ratio of the radii of their circular paths?

A proton, a deuteron and an $α$ -particle are projected perpendicular to the direction of a uniform magnetic field with same kinetic energy. The ratio of the radii of the circular paths described by them is

Two particles X and Y having equal charges after being accelerated through same potential difference enter a region of uniform magnetic field and describe a circular paths of radii ' $r_{1}$ and ' $r_{2}$ ' respectively. The ratio of the mass of X to that of Y is

A proton and an $α$ -particle, moving with the same velocity, enter a uniform magnetic field, acting normal to the plane of their motion. The ratio of the radii of the circular paths described by the proton and $α$ -particle is

A particle of mass m, charge $q$ and kinetic energy $t$ enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

A proton, a deuteron and an $α - p a r t i c l e$ having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_{p}, r_{d}$ and $r_{α}$ denote, respectively, the radii of the trajectories of these particles, then