Use app

Motion in the Presence of Magnetic Field - Problems L1

Q: 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°\$

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 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\$

Q: 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

Q: 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.

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\$

Physics

formula

Radius of curvature of a particle moving in a magnetic field

r = \frac{m v}{q B}

REVISE WITH CONCEPTS