Question

The torque on a current-carrying loop in a uniform magnetic field is maximum when:

पर्याय

• The plane of the loop is parallel to \[\vec{B}\] (θ = 90°)

• The plane of the loop is perpendicular to \[\vec{B}\] (θ = 0°)

• The number of turns N is reduced to zero

• The plane of the loop makes an angle of 45° with \[\vec{B}\]

Answer 1

Since \[\tau = NIAB\sin\theta\], the torque is maximum when \[\sin\theta = 1\], i.e., θ = 90°. This corresponds to the plane of the loop being parallel to \[\vec{B}\], giving \[\tau_{max} = NIAB\].