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Moving Charges and Magnetism Test - 70

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Moving Charges and Magnetism Test - 70
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  • Question 1
    1 / -0
    In case Hall effect for a strip having charge $$Q$$ and area of cross-section $$A$$, the Lorentz force is
    Solution
    $$\vec F = Q[\vec E + (\vec v \times \vec B)]$$
    So, $$F \propto Q$$
  • Question 2
    1 / -0
    " On flowing current in a conducting wire the magnetic field produces around it". It is a law of 
    Solution
    The aboove statement is ampere's law.
  • Question 3
    1 / -0
    Mixed $$He^+$$ and $$O^{2+}$$ ions (mass of $$He^+=4$$ amu and that of $$O^{2+}=16$$ amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then
    Solution
    $$r=\dfrac {\sqrt {2mK}}{qB}\Rightarrow r\propto \dfrac {\sqrt m}{q}\Rightarrow \dfrac {r_{He^+}}{r_{O^{++}}}=\sqrt {\dfrac {m_{He^+}}{m_{O^{++}}}}\times \dfrac {q_{O^{++}}}{q_{He^+}}$$
    $$=\sqrt {\dfrac {4}{16}}\times \dfrac {2}{1}=\dfrac {1}{1}$$. Then will deflect equality.
  • Question 4
    1 / -0
    If cathode rays are projected at right angles to a magnetic field, their trajectory is
    Solution

    Let the magnetic field be along North direction. Let cathode ray pass perpendicular to the magnetic field towards east.then it gets deflected downwards.
    Fleming's left hand rule is meant for the movement of positively charged particles. But cathode rays are stream of negatively charged electrons. So opposite to the previous one.
    And like postive charge it will make a circular trajectory.
  • Question 5
    1 / -0
    A beam of electrons and protons move parallel to each other in the same direction, then they
    Solution
    The flow of positive chage is taken as the direction of current. 
    So, here the currents are in opposite direction so, they will repel each other.
  • Question 6
    1 / -0
    Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a $$25\ cm$$ length of wire $$C$$ is

    Solution
    Force on wire $$C$$ due to wire $$D$$
    $$F_D=10^{-7} \times \dfrac{2 \times 30 \times 10}{2 \times 10^{-2}} \times 25 \times 10^{-2}=5 \times 10^{-4}\ N$$ (towards right)
    Force on wire $$C$$ due to wire $$G$$

    $$F_G=10^{-7} \times \dfrac{2 \times 20 \times 10}{2 \times 10^{-2}} \times 25 \times 10^{-2}=5 \times 10^{-4}\ N$$ (towards left)
    $$\Rightarrow$$ Net force on wire $$C$$ is $$F_{net}=F_D- F_G=0$$

  • Question 7
    1 / -0
    An electron and a proton have equal kinetic energies. They enter in a magnetic perpendicular. They 
    Solution
    Particles entering perpendicular, hence they will describe circular path. Since their masses are different so they will described path of difference radii.
  • Question 8
    1 / -0
    Assertion : An electron is not deflected on passing through certain region of space. This observation confirms that there is no magnetic field in that region.
    Reason : The deflection of electron depends on angle between velocity of electron and direction of magnetic field.
    Solution
    If electrons is moving parallel to the magnetic field, then the electron is not deflected i.e., if the electron is not deflected we cannot be sure that there is no magnetic field in that region.
  • Question 9
    1 / -0
    The distance between the supply wires of an electric-mains is $$12cm$$. These wires experience $$4mg$$ weight per unit length. Current flowing in both the wire will be if they carry current in same direction.
    Solution

  • Question 10
    1 / -0
    A long, vertical, metallic wire carries downward electric current.
    What would be the direction of the field if the current consisted of positive charges moving downward instead of electrons moving upward? 
    Solution
    The direction of the magnetic field at a given point is determined by the direction of the conventional current that creates it.
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