Q: 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\$.

Q: 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}\$

Q: 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\$

Q: 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\$.

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\$

Question 1

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:-

Accepted Answer

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 $90&#176;$

Question 2

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?

Accepted Answer

$\dfrac {mv^{2}}{R} = qvB$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$.

Question 3

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?

Accepted Answer

Toppr Ask Answer

Question 4

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

Accepted Answer

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$ $r_p : r_d:r_{\alpha} =\sqrt2:1:\sqrt{2}$

Question 5

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

Accepted Answer

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 $Y$ $\dfrac{m_1}{m_2} = \bigg[\dfrac{r_1}{r_2} \bigg]^2$

Question 6

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

Accepted Answer

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$

Question 7

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

Accepted Answer

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

Question 8

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:

Accepted Answer

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.

Question 9

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

Accepted Answer

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$.

Question 10

An electron is moving with a speed of $3 \times 1 0^{- 7} m / s$ in a magnetic field of $6 \times 1 0^{- 4} T$ 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 $= 9 \times 1 0^{- 31} k g, c h a r g e e = 1.6 \times 1 0^{- 19} C, e V = 1.6 \times 1 0^{- 19} J$ ]

Accepted Answer

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$

Charged Particle in Uniform Magnetic Field - II

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REVISE WITH CONCEPTS

Motion in the Presence of Magnetic Field

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

Charged Particle in Uniform Magnetic Field - II

Language

Rate

REVISE WITH CONCEPTS

Motion in the Presence of Magnetic Field

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