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Moving Charges and Magnetism Test - 61

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Moving Charges and Magnetism Test - 61
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  • Question 1
    1 / -0
    The force experienced by a pole of strength $$100$$ Am at a distance of $$0.2$$ m from a short magnet of length $$5 m$$ and pole strength of $$200$$ Am on its axial line will be:-
    Solution

  • Question 2
    1 / -0
    Following figure shows the path of an electron that passes through two regions containing uniform magnetic fields of magnitudes $$B_1$$ and $$B_2$$. Its path in each region is a half circle,choose the correct option

    Solution
    Equating centripetal force with magnetic force,
    $$\dfrac{mv^2}r = qvB$$
    $$r = \dfrac{mv}{qB}$$
    Thus path having greater r will have lower magnetic field. Therefore field $$B_1$$ is stronger.
    Considering the right hand thumb rule, the direction of the magnetic field for region of magnetic field $$B_1$$ is into the page and where for region $$B_2$$ is out of the page.
  • Question 3
    1 / -0
    $$AB$$ and $$CD$$ are long straight conductor, distance $$d$$ apart, carrying a current $$I$$. The magnetic field at the midpoint of $$BC$$ is

    Solution
    The field at the midpoint of $$BC$$ due to $$AB$$ is $$\left(-\dfrac{\mu_{0}}{4\pi}.\dfrac{i}{d/2}\hat{k}\right)$$ and the same is due to $$CD$$. 

    Therefore the total field is $$\left[-\left(\dfrac{\mu_{0}i}{\pi d}\right)\hat{k}\right]$$
  • Question 4
    1 / -0
    A magnetic filed:
    Solution

  • Question 5
    1 / -0

    Magnetic field due to a horseshoe magnet is parallel to the plane of the paper and North Pole is on the left. Direct current in a straight wire is perpendicular to the plane of the paper, toward the observer. The wire will be deflected

    Solution
    Magnetic field due to horse shoe magnet is parallel to the plane of the paper and North pole is on the left.
    The wire will be deflected on the right due to I and moving wire will be $$180°$$ to each other.

  • Question 6
    1 / -0
    An electron has velocity v=$$\left( {2.0\times{{10}^6}\;m/s} \right)\hat{i} + \left( {3.0\times{{10}^{6\;}}m/s} \right)\hat{j}$$, magnetic field present is region is $$B\left( {0.030\;T} \right)\;i - \left( {0.15\;T} \right)j$$. Find the force on electron.
    Solution

  • Question 7
    1 / -0
    A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field $$\overline {B}$$ at a point having coordinates (x, y) in the z=0 plane is :
    Solution
    An infinite wire carrying current in $$(-z)$$ axis
    we have $$\vec { l } =\left( -\hat { k }  \right) $$ (current vector)
    $$\vec { l } =x\hat { i } +y\hat { j } \Rightarrow { r }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$$
    using Biot-Savart law fo rinfinite wire
    $$\vec { B } =\cfrac { { \mu  }_{ 0 }I }{ 2\pi r } \left( \cfrac { \vec { l } \times \vec { r }  }{ r }  \right) $$ ($$\cfrac { \vec { l } \times \vec { r }  }{ r }  $$ is the unit vector along $$\vec { B } $$)
    $$=\cfrac { { \mu  }_{ 0 }I }{ 2\pi { r }^{ 2 } } \left[ \left( -\hat { k }  \right) \times \left( x\hat { i } +y\hat { j }  \right)  \right] \Rightarrow \vec { B } =\cfrac { { \mu  }_{ 0 }I }{ 2\pi ({ x }^{ 2 }+{ y }^{ 2 }) } \left( y\hat { i } -x\hat { j }  \right) $$ (Ans)

  • Question 8
    1 / -0
    Two free parallel straight wires carrying currents in opposite direction
    Solution
    Two wires, if carries current in opposite direction, they repel each other.
  • Question 9
    1 / -0
    In the given region, magnetic field is applied in the direction perpendicular to the plane of paper and outward. If the charge particle goes undeflected, then the direction electric field present in the region is- 

  • Question 10
    1 / -0
    A charged particle moving in a magnetic field experience maximum magnetic force when 
    Solution

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