Self Studies

Moving Charges and Magnetism Test - 41

Result Self Studies

Moving Charges and Magnetism Test - 41
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    An $$\alpha$$ -particles enters at the middle as shown in Fig. with $$10^5 ms^{-1}$$. In which direction will it bend?

    Solution
    According to magnetism concepts of straight wire 3A wire will produce stronger magnetic field inside the page and 1A wire will produce 1/3rd of it out of the page. Thus net magnetic field is inward. And the magnetic force on charge is the vector product of velocity and magnetic field. Thus net force will be towards 3 A wire.
  • Question 2
    1 / -0
    The phenomenon of production of magnetic field on passing an electric current in a straight conducting wire is based on the law of
    Solution
    The phenomenon of production of magnetic field on passing an electric current in a straight conducting wire is based on the law of Ampere.
  • Question 3
    1 / -0
    A straight conductor carries a current. Assume that all free electrons in the conductor move with the same drift velocity v. A and B are two observers on a straight line XY parallel to the conductor. A is stationary. B moves along XY with a velocity v in the direction of the free electrons.
    Solution
    A is stationary and observes the current I. B observes the free electrons to be at rest, but the unbalanced positive charges in the conductor will appear to move in the direction opposite to that of v. Thus A and B observe the same current and hence same magnetic field. 
  • Question 4
    1 / -0
    A conducting gas is in the form of a long cylinder. Current flows through the gas along the length of the cylinder. The current is distributed uniformly across the cross-section of the gas. Disregard thermal and electrostatic forces among the gas molecules. Due to the magnetic fields set up inside the gas and the forces which they exert on the moving ions, the gas will tend to
    Solution
    Two parallel current carrying conductors attract each other.
    The gas will contract since it can be considered as a no of  parallel streams of current in the same direction.
    Hence, there will be contraction.
  • Question 5
    1 / -0
    Two parallel straight conductors, in which current is flowing in the same direction, attract each other. The cause of it is
    Solution

    The force between two long parallel conductors separated by distance r, is $$F =\dfrac{\mu_0 (i_1)(i_2)}{2 \pi r}$$

    If the current is flowing in the same direction through both conductors, they will attract each other due to magnetic force between the two.

  • Question 6
    1 / -0
    The work done by a normal magnetic field in revolving a charged particle q in a circular path will be
    Solution
    $$\vec{F_{magnetic}} = q(\vec{v}\times \vec{B})$$
    $$ \therefore \vec{F_{magnetic}}.\vec{v} = 0$$
    Hence, there is no power supplied to the charged particle over a cycle.
    $$ \Rightarrow$$ work done is zero. 
  • Question 7
    1 / -0
    A charged particle with charge $$q$$ is moving in a uniform magnetic field. If this particle makes any angle with the magnetic field then its path will be :
    Solution
    The component of velocity( $$ v_{\parallel}$$) parallel to magnetic field $$\vec{B}$$ will make the charged particle move parallel or anti parallel  to $$\vec{B}$$
    The $$ v_{\perp}$$ will make the particle move in a circle, plane of the circle being  perpendicular to the direction of the magnetic field.
  • Question 8
    1 / -0
    The distance between two thin long straight parallel conducting wires is $$b$$. On passing the same current $$i$$ in them, the force per unit length between them will be
    Solution
    The force between two long parallel conductors separated by distance r, is
    $$F = \dfrac{\mu_0 (i_1)(i_2)}{2 \pi r}$$
    Here, both conductors carry same current and they are separated by distance b, hence the force between them is
    $$F = \dfrac{\mu_0 (i)^2}{2 \pi b}$$
  • Question 9
    1 / -0
    If the currents in two straight current-carrying conductor, distant d apart, are $$i_1$$ and $$i_2$$ respectively in the same direction then they will
    Solution
    Force experienced by one current carrying conductor due to another parallel straight conductor per unit length is,
    $$F = \dfrac{\mu_0 i_1 i_2}{2\pi r}$$
    where, variables have their usual meanings.
    If the current flowing through the conductors is in the same direction they will attract each other. If current flows in opposite direction through both the conductors then they will repel each other.
  • Question 10
    1 / -0
    The rays, which remain undeflected in a magnetic field, are:
    Solution
    $$\gamma$$ particles do not have any charge on it, hence $$\gamma$$-rays remains undeflected in magnetic field.
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now