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This is one of our many quizzes on Lorentz force. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Lorentz force Quiz 2

physics

 The magnetic flux linked with a solenoid is $4\times {{10}^{-3}}\text{ Wb}$ while the area of cross-section of the solenoid is $10\text{ c}{{\text{m}}^{2}}$. A charged ion of specific charge ${{10}^{10}}\text{ C k}{{\text{g}}^{-1}}$ is projected normally to the field inside the solenoid with a velocity $2\times {{10}^{7}}\text{ m}{{\text{s}}^{-1}}$. What is the radius of the circular trajectory followed by the ion? 

 Calculate magnetic field from flux and substitute it in the expression for Lorentz force experienced by the ion. 

  We know that magnetic flux is given by $ \\\phi =BA \\\Rightarrow B=\dfrac{\phi }{A} \\\Rightarrow B=\dfrac{4\times {{10}^{-3}}}{10\times {{10}^{-4}}} \\\Rightarrow B=4\text{ T }\to \text{(1)} \\ $ Now, when the ion is projected normally to this magnetic field linked with the solenoid, we have: $ \\{{F}_{L}}={{F}_{C}} \\\dfrac{m{{v}^{2}}}{r}=qvB\text{ }\to \text{(2)} \\ $ From the question, we are provided that specific charge of the ion, $ \\\dfrac{q}{m}={{10}^{10}}\text{ C k}{{\text{g}}^{-1}}\text{}\to \text{(3)}\\ $ Substituting $(1)$ and $(3)$ in $(2)$, we have $ \\r=\dfrac{mv}{qB} \\\Rightarrow r=\dfrac{2\times {{10}^{7}}}{{{10}^{10}}\times 4} \\\therefore r=0.5\times {{10}^{-3}}\text{ m}$ 

 The frequency of a charged particle used in a cyclotron is equal to $250\text{ Hz}$. What is the velocity of this particle if it is moving perpendicular to the uniform magnetic field applied to the particle?   $($Take charge of the particle to be $1.2\times {{10}^{-4}}\text{ C}$ and Lorentz force on the particle to be $\text{1}\text{.88 N})$ 

 Recall the expressions for Lorentz force and cyclotron frequency and combine both of them to arrive at the required velocity. 

  We are given that cyclotron frequency, $ \\f=250\text{ Hz}$ said that the charged particle is moving perpendicular to the applied magnetic field, from which, $\\\sin \theta =\sin 90{}^\circ =1\text{}\to \text{(1)}\\ $ We know that cyclotron frequency is given by $ \\f=\dfrac{Bq}{2\pi m} \\\Rightarrow Bq=2\pi fm \\\Rightarrow Bq=2\times 3.14\times 250\times 1.2\times {{10}^{-4}} \\\Rightarrow \left| Bq \right|=0.18\text{ }\to \text{(2)} \\ $ Lorentz force on a charged particle moving in a uniform magnetic field is given by $ {{F}_{L}}=Bqv\sin \theta$ $\\ $ Substituting given value of Lorentz force,$(1)$ and $(2)$ in the above expression, we have $\\1.88=1.88\times {{10}^{-4}}\times v\times 1 \\\therefore v={{10}^{4}}\text{ m}{{\text{s}}^{-1}}$

 The momentum of an ion moving at right angle to a magnetic field $'x'$ is $0.86\text{ kgm}{{\text{s}}^{-1}}$. If radius of the trajectory followed by the ion is $0.2\text{ cm}$, what is the value of $'x'$? $($Take the charge of ion to be $2\times {{10}^{-12}}\text{ C})$

 Combine momentum, Lorentz force and centripetal force to arrive at the answer. 

 In this case, Lorentz force acting on the ion can be equated to the centripetal force acting on it.Thus, $ \\{{F}_{L}}={{F}_{C}} \\\Rightarrow qvB\sin \theta =\dfrac{m{{v}^{2}}}{r} \\\Rightarrow qvx=\dfrac{m{{v}^{2}}}{r}\text{    }\left( \because \sin \theta =\sin 90{}^\circ =1 \right) \\\Rightarrow x=\dfrac{mv}{qr} \\\Rightarrow x=\dfrac{p}{qr}\text{     }\left( \because p=mv \right) \\\Rightarrow x=\dfrac{0.86}{2\times {{10}^{-12}}\times 0.2\times {{10}^{-2}}} \\\Rightarrow x=\dfrac{0.86}{4\times {{10}^{-15}}} \\\therefore x=2.15\times {{10}^{16}}\text{ T}$

 Evaluate the diameter of the circular trajectory followed by a positron moving with an angular deviation of $45{}^\circ $ to a field of $0.8\text{ T}$ and with a velocity of $3\times {{10}^{2}}\text{ m}{{\text{s}}^{-1}}$.  

 Recognise the equality in forces acting on the positron in the given atmosphere and take care of the substitutions.  

 We know that mass and charge of a positron are $ 9.1\times {{10}^{-31}}\text{ kg}$ and $1.6\times {{10}^{-19}}\text{ C}$ respectively. $ \\ $ When a positron is moving in a uniform magnetic field with an angular deviation of $45{}^\circ$ $ \\ $ Lorentz force on the positron can be written as $ \\{{F}_{L}}=qvB\sin \theta  \\\Rightarrow {{F}_{L}}=qvB\sin (45{}^\circ ) \\\Rightarrow {{F}_{L}}=\dfrac{qvB}{\sqrt{2}} \\ $ Now, since the positron is moving in a circular path, the centripetal force experienced by the positron can be equated to the Lorentz force on it as follows: $ \\{{F}_{L}}=\dfrac{m{{v}^{2}}}{r} \\\Rightarrow \dfrac{qvB}{\sqrt{2}}=\dfrac{m{{v}^{2}}}{r} \\\Rightarrow qBr=\sqrt{2}mv \\\Rightarrow 1.6\times {{10}^{-19}}\times 0.8\times r=\sqrt{2}\times 9.1\times {{10}^{-31}}\times 3\times {{10}^{2}} \\\Rightarrow r=\dfrac{38.6\times {{10}^{-29}}}{1.28\times {{10}^{-19}}} \\\Rightarrow r=30.15\times {{10}^{-10}}\text{ m} \\\Rightarrow r=30.15\overset{\text{o}}{\mathop{\text{A}}} \\\Rightarrow d=2r \\\therefore d=60.3\overset{\text{o}}{\mathop{\text{A}}}$ 

 Choose the correct statement: If the ratio of perimeters of circular trajectories of two charged particles $'x'$ and $'y'$ in an electromagnetic atmosphere is $2$, then their, $ \\ $ $(A)$ Speeds are equal $ \\ $ $(B)$ Momentums are equal $ \\ $ $(C)$ Kinetic energies are equal $ \\ $ $(D)$ Speeds as well as their momentums are equal $\left(\text{Given }\dfrac{{{m}_{x}}}{{{m}_{y}}}=4\text{ and }\dfrac{{{q}_{x}}}{{{q}_{y}}}=2\right)$

 Obtain an expression for radius of the circular trajectory for a charged particle in an electromagnetic atmosphere by considering the forces acting on it and utilize the same to relate it with the given ratio of perimeters of $'x'$ and $'y'$. 

 Let the perimeters of $x$ and $y$ be $2\pi {{r}_{x}}$ and $2\pi {{r}_{y}}$ respectively. $ \\ $ Given that:$ \\\dfrac{2\pi {{r}_{x}}}{2\pi {{r}_{y}}}=2\text{} \\\Rightarrow \dfrac{{{r}_{x}}}{{{r}_{y}}}=2\text{   }\to \text{(1)} \\ $ For a charged particle entering an electromagnetic atmosphere with its velocity perpendicular to the magnetic field, we know that $ \\{{F}_{L}}={{F}_{C}} \\\Rightarrow qvB=\dfrac{m{{v}^{2}}}{r} \\\Rightarrow r=\dfrac{mv}{qB}\text{     }\to \text{(2)} \\ $ From $(1)$ and $(2)$, we can deduce that: $ \\\dfrac{{{r}_{x}}}{{{r}_{y}}}=\dfrac{{{m}_{x}}{{v}_{x}}}{{{q}_{x}}B}\times \dfrac{{{q}_{y}}B}{{{m}_{y}}{{v}_{y}}} \\\Rightarrow 2=\dfrac{{{m}_{x}}{{v}_{x}}}{{{q}_{x}}B}\times \dfrac{{{q}_{y}}B}{{{m}_{y}}{{v}_{y}}} \\\Rightarrow 2=\dfrac{{{m}_{x}}}{{{m}_{y}}}\times \dfrac{{{v}_{x}}}{{{v}_{y}}}\times \dfrac{{{q}_{y}}}{{{q}_{x}}} \\\Rightarrow 2=4\times \dfrac{{{v}_{x}}}{{{v}_{y}}}\times \dfrac{1}{2} \\\Rightarrow 2=2\times \dfrac{{{v}_{x}}}{{{v}_{y}}} \\\therefore {{v}_{x}}={{v}_{y}} \\\text{Also, since}\,{{m}_{x}}\ne {{m}_{y}}\,$ which are directly proportional to momentum and kinetic energy, we can conclude that only their speeds are equal.