The torque on a current loop is zero if the angle between the positive normal and the magnetic field is either `theta =0` or `theta = 180^@`. In which of the two orientations, the equilibrim is stable?

The force exerted on a current-carrying wire placed in magnetic field is zero when the angle between the wire and the direction of magnetic field is :

If there is uniform magnetic field of in the positive z direction , in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ?

A rectangular loop of sides 10 cm and 5 cm carrying a current 1 of 12 A is placed in different orientations as shown in the figures below : ltBrgt If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium ?

A rectangular loop of sides 10cm and 5cm carrying a current I of 12A is placed in different orienctations as shown in the figure If there is a uniform magnetic field of 0.3 T in the positive z-direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium .

The path of a charged particle in a uniform magnetic field depends on the angle theta between velocity vector and magnetic field, When theta is 0^(@) or 180^(@), F_(m) = 0 hence path of a charged particle will be linear. When theta = 90^(@) , the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius r = (mv)/(qB) . The time period for circular path will be T = (2pim)/(qB) When theta is other than 0^(@), 180^(@) and 90^(@) , velocity can be resolved into two components, one along vec(B) and perpendicular to B. v_(|/|)=cos theta v_(^)= v sin theta The v_(_|_) component gives circular path and v_(|/|) givestraingt line path. The resultant path is a helical path. The radius of helical path r=(mv sin theta)/(qB) ich of helix is defined as P=v_(|/|)T P=(2 i mv cos theta) p=(2 pi mv cos theta)/(qB) Two ions having masses in the ratio 1:1 and charges 1:2 are projected from same point into a uniform magnetic field with speed in the ratio 2:3 perpendicular to field. The ratio of radii of circle along which the two particles move is :

The path of a charged particle in a uniform magnetic field depends on the angle theta between velocity vector and magnetic field, When theta is 0^(@) or 180^(@), F_(m) = 0 hence path of a charged particle will be linear. When theta = 90^(@) , the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius r = (mv)/(qB) . The time period for circular path will be T = (2pim)/(qB) When theta is other than 0^(@), 180^(@) and 90^(@) , velocity can be resolved into two components, one along vec(B) and perpendicular to B. v_(|/|)=cos theta v_(^)= v sin theta The v_(_|_) component gives circular path and v_(|/|) givestraingt line path. The resultant path is a helical path. The radius of helical path r=(mv sin theta)/(qB) ich of helix is defined as P=v_(|/|)T P=(2 i mv cos theta) p=(2 pi mv cos theta)/(qB) A charged particle moves in a uniform magnetic field. The velocity of particle at some instant makes acute angle with magnetic field. The path of the particle will be

The path of a charged particle in a uniform magnetic field depends on the angle theta between velocity vector and magnetic field, When theta is 0^(@) or 180^(@), F_(m) = 0 hence path of a charged particle will be linear. When theta = 90^(@) , the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius r = (mv)/(qB) . The time period for circular path will be T = (2pim)/(qB) When theta is other than 0^(@), 180^(@) and 90^(@) , velocity can be resolved into two components, one along vec(B) and perpendicular to B. v_(|/|)=cos theta v_(^)= v sin theta The v_(_|_) component gives circular path and v_(|/|) givestraingt line path. The resultant path is a helical path. The radius of helical path r=(mv sin theta)/(qB) ich of helix is defined as P=v_(|/|)T P=(2 i mv cos theta) p=(2 pi mv cos theta)/(qB) Which particle will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field.

A bar magnet of magnetic moment 6 J//T is aligned at 60^(@) with a uniform external magnetic field of 0.44 T. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field, (ii) opposite to the magnetic field, and (b) the torque on the magnet in the final orientation in case (ii).

Force on current carrying loop (Radius = R) in uniform magnetic (B) field which is at an angle 30” with the normal will be :-

Mention the angle between a current carrying conductor and magnetic field for which the force experienced by this current conductor placed field is largest ?