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Moving Charges and Magnetism Test - 22

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Moving Charges and Magnetism Test - 22
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  • Question 1
    1 / -0
    Two parallel beams of positrons moving in the same direction will
    Solution
    Two parallel beams of positron moving in same direction is equivalent to two current carrying conductor, carrying current in same direction. 
    Hence, they will attract each other.
  • Question 2
    1 / -0
    An uncharged particle is moving with a velocity of $$\vec { v } $$ in non-uniform magnetic field as shown. Velocity $$\vec { v } $$ would be:

    Solution
    Magnetic force            $$\vec{F_m} = q (\vec{v} \times \vec{B})$$
    The particle being uncharged, $$q=0$$.
    So, magnetic force on the particle, $$\vec{F_m}=\vec{0}$$
    Since there is no external force acting on the particle, its velocity $$\vec{v}$$ will be same at all points.
  • Question 3
    1 / -0
    When a positively charged particle enters into a uniform magnetic field with uniform velocity, its trajectory can be
    (i) A straight line
    (ii) A circle
    (iii) A helix
    Solution
    Force on a moving charge in a magnetic field is $$q(\vec{v}\times \vec{B})$$.
    Thus if the particle is moving along the magnetic field, $$\vec{F}=0$$.
    Hence the particle continues to move along the incident direction, in a straight line.
    When the particle is moving perpendicular to the direction  of magnetic field, the force is perpendicular to both direction of velocity and the magnetic field.
    Then the force tends to move the charged particle in a plane perpendicular to the direction of magnetic field, in a circle.
    If the direction of velocity has both parallel and perpendicular components to the direction magnetic field, the perpendicular component tends to move it in a circle and parallel component tends to move it along the direction of magnetic field. Hence the trajectory is a helix.
  • Question 4
    1 / -0
    A charged particle enters a magnetic field H with its initial velocity making an angle of $$45^o$$ with H. Then the path of the particle will be
    Solution
    At point A, charge is entering in a magnetic field in which direction of field is shown in the figure. The velocity of particle is u making an angle of $$45^o$$ with field. We resolve it in two directions, one along the field and other perpendicular to it. Since u $$\sin 45^o$$ is perpendicular to H, it will create a rotatory effect on the charge. So charge particle will start rotating with axis along the direction of H. At the same time it will move forward with velocity u $$\cos 45^o$$. Under both these motions, it will have helical path as shown in the figure. 
  • Question 5
    1 / -0
    Two wires carrying
    Solution
    The direction of magnetic field can be determined from right hand thumb rule. The direction of magnetic force is given by the cross product $$ \vec { F } =i(\vec { l } \times \vec { B } ) $$.
    In the case of two wires carrying parallel current, this force on one wire due to the magnetic field of the other is directed towards the other wire. So, wires carrying parallel current attract each other.
    In the case of two wires carrying anti-parallel current, this force on one wire due to the magnetic field of the other is directed away from the other wire. So, wires carrying anti-parallel current repel each other.

  • Question 6
    1 / -0
    An electron is traveling along the x-direction. It encounters a magnetic field in the y-direction. Its subsequent motion will be
    Solution
    When a charged particle enters a magnetic field, it experiences a force which is always directed perpendicular to its direction of motion. In that case the path of the charged particle becomes circular and the plane of the circle is perpendicular to the plane containing the magnetic field and velocity vector of the charged particle.
  • Question 7
    1 / -0
    A deuteron of kinetic energy $$50\ keV$$ is describing a circular orbit of radius $$0.5$$ metre in a plane perpendicular to the magnetic field $$B$$. The kinetic energy of the proton that describes a circular orbit of radius $$0.5$$ metre in the same plane with the same $$B$$ is
    Solution
    For a charged particle orbiting in a circular path in a magnetic field
    $$\dfrac {mv^{2}}{r} = Bqv\Rightarrow v = \dfrac {Bqr}{m}$$
    or, $$mv^{2} = Bqvr$$
    Also,
    $$E_{K} = \dfrac {1}{2}mv^{2} = \dfrac {1}{2}Bqvr = Bq \dfrac {r}{2}. \dfrac {Bqr}{m} = \dfrac {B^{2}q^{2}r^{2}}{2m}$$
    For deuteron, $$E_{1} = \dfrac {B^{2}q^{2}r^{2}}{2\times 2m}$$
    For proton, $$E_{2} = \dfrac {B^{2}q^{2}r^{2}}{2m}$$
    $$\dfrac {E_{1}}{E_{2}} = \dfrac {1}{2}\Rightarrow \dfrac {50keV}{E_{2}} = \dfrac {1}{2} \Rightarrow E_{2} = 100 keV$$.
  • Question 8
    1 / -0
    When $$0.50\overset {\circ}{A}$$ X-rays strike a material, the photoelectrons from the $$k$$ shell are observed to move in a circle of radius $$23\ mm$$ in a magnetic field of $$2\times 10^{-2}$$ tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of $$k-shell$$ electrons?
    Solution
    As we know,
    $$F = qvB = m \dfrac {v^{2}}{R} \Rightarrow v = \dfrac {q}{m}BR$$
    The kinetic energy of the photoelectron
    $$\dfrac {1}{2} mv^{2} = \dfrac {1}{2} \dfrac {e^{2}B^{2}R^{2}}{m}$$
    $$= \dfrac {1}{2} \dfrac {(1.6\times 10^{-19})^{2} (2\times 10^{-2})^{2} (23\times 10^{-3})^{2}}{(9.1\times 10^{-31})}$$
    $$= 2.97\times 10^{-15}J$$
    $$= \dfrac {2.97\times 10^{-15}}{1.6\times 10^{-19}} = 18.36\ keV$$
    Energy of the incident photon $$= \dfrac {hc}{\lambda}$$
    $$= \dfrac {12.4}{0.50} = 24.8\ keV$$
    Therefore, Binding energy $$= 24.8 - 18.6 = 6.2\ keV$$.
  • Question 9
    1 / -0
    A positively charged particle projected towards east is deflected towards north by a magnetic field. The field may be :
    Solution
    The direction of the positive charge particle towards east and hence the current's direction and the direction of the force towards north. Now by applying Fleming left hand rule we can say that the direction of the field towards downward. 
  • Question 10
    1 / -0
    When a charged particle is acted on only by a magnetic force, its:
    Solution
    Magnetic force acting on a charge $$q$$ travelling with velocity $$v$$ is given by $$\vec{F}=q(\vec{v}\times \vec{B})$$
    Since the force is perpendicular to the velocity, work done by force on the object is always zero. 
    Hence the total energy of the charged particle does not change.
    Hence correct answer is option D.
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