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Moving Charges and Magnetism Test - 82

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Moving Charges and Magnetism Test - 82
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  • Question 1
    1 / -0
    Two identical particles having the same mass m and charges +q and - q separated by a distance d enter in a uniform magnetic field B directed perpendicular to paper inwards with speed $$v _ { 1 } \text { and } v _ { 2 }$$ as shown in figure. The particles will not collide if

    Solution
    Given : Two identical particles having the same mass m and charges +q and - q separated by a distance d enter in a uniform magnetic field B directed perpendicular to paper inwards with speed V1 and V2.

    Solution :
    Formula for radius of circular path of particle due moving inside magnetic field is $$R=\frac{mv}{qB}$$
                      $$R_{1}=\frac{mv_{1}}{qB}$$        ,      $$R_{2}=\frac{mv_{2}}{qB}$$

    Force on moving charge $$F=q(\vec{V} \times \vec{B})$$
    From this we can tell direction of force on the particles 
    Particle with negative charge will move downwards and with positive charge will move upward in the given case.

    For particles to not collide d > $$R_{1}+R_{2}$$

                                                d > $$\frac{mv_{1}}{qB}$$  +  $$\frac{mv_{2}}{qB}$$

                                                d >$$\frac{m}{qB} (v1+v2)$$

    The correct opt :A

  • Question 2
    1 / -0
    Proton and $$\alpha$$ particle enters in stable uniform magnetic field with same kinetic energy. If the radius of circular path are $$r_{p}$$ and $$r_{a}$$, then,
    Solution

  • Question 3
    1 / -0

    A particle with charge q and , mass m starts its motion from origin with a velocity $$\overrightarrow v  = a\widehat i$$. A uniform magnetic field $$\overrightarrow B  = \dfrac{b}{4}\widehat i + \dfrac{{\sqrt 3 b}}{4}\widehat j$$ exists everywhere in space. The component of the velocity vector in y-direction when the value of z-coordination becomes maximum is 

  • Question 4
    1 / -0
    Several $$\alpha-$$ particle of different speeds enter a uniform magnetic field confined into a cylindrical region. if all the $$\alpha-$$ particles enter the field radially, what can say about time intervals spent by them in the magnetic field?
  • Question 5
    1 / -0
    Two point charges $$A$$ and $$B$$ of same charge having magnitude of momentum $$p_1$$ and $$p_2$$ respectively and having same charge are moving in a plane containing uniform magnetic field perpendicular to the plane. Then (Trajectories as shown in figure)  

    Solution

  • Question 6
    1 / -0
    Current $$I$$ flowing along edges of face of a cube as shown in the figure $$i$$, produces magnetic field $$\vec{B}=B_{o}\hat{j}$$ at the centre of the cube. Consider another identical cube. where the current. $$I$$ flows along the path shown in the figure $$II$$. How much magnetic field exist at the centre of the second cube?

    Solution

  • Question 7
    1 / -0
    A $$\gamma -$$ray photon is passing near a nucleus, and breaks into an electron and positron. The region contains a uniform magnetic field $$B$$ perpendicular to the plane of motion. The time after which they again converted into $$\gamma -$$ray is : 
    [The force of electrostatic interaction and gravitational interaction may be neglected]
    Solution

  • Question 8
    1 / -0
    A particle of mass $$m$$, carrying in a uniform magnetic field directed along $$x-$$axis. At the instant $$t=0$$ it is given a velocity $$v_0$$ at an angle $$\theta$$ with the $$y-$$axis, in the $$x-y$$ plane. The coordinates of the particle after one revolution will be :

    Solution

  • Question 9
    1 / -0
    A long straight wire and a circular loop carrying currents $$I_1$$ and $$I_2$$ respectively are in the same plane as shown in figure. If the magnetic field at the centre of the loop is zero, then :

    Solution

  • Question 10
    1 / -0
    An electron accelerated through a potential difference $$V$$ enters into a uniform transverse magnetic field and experiences a force $$F$$. If the acceleration potential is increased to $$2V$$, the electron in the same magnetic field will experience a force :
    Solution

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