$A$ current-carrying loop is free to turn in a uniform magnetic field. The loop will come into equilibrium when its plane is inclined at

  • A
    $0^\circ$ to the direction of the field
  • B
    $45^\circ$ to the direction of the field
  • C
    $90^\circ$ to the direction of the field
  • D
    $135^\circ$ to the direction of the field

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$A$ circular coil carrying a current of $2.5 \ A$ is free to rotate about an axis in its plane perpendicular to an external magnetic field. When the coil is made to oscillate,the time period of oscillation is $T$. If the current through the coil is $10 \ A$,the time period of oscillation is

$A$ uniform magnetic field of $3000\,G$ is established in the positive $z-$ direction. $A$ rectangular loop of sides $10\,cm$ and $5\,cm$ carries a current of $12\,A$. The loop is placed in the $xy-$ plane as shown in the figure. What is the magnitude of the torque on the loop?

$A$ circular coil of $16$ turns and radius $10 \;cm$ carrying a current of $0.75 \;A$ rests with its plane normal to an external field of magnitude $5.0 \times 10^{-2} \;T$. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released,it oscillates about its stable equilibrium with a frequency of $2.0 \;s^{-1}$. What is the moment of inertia of the coil about its axis of rotation?

$(a)$ $A$ current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e.,turns about the vertical axis)?
$(b)$ $A$ current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn,what is its orientation of stable equilibrium? Show that in this orientation,the flux of the total field (external field $+$ field produced by the loop) is maximum.
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