A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

A plane is inclined at an angle $$\alpha $$ = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms^–1 , from the base of the plane, making an angle $$\theta $$ = 15° with respect to the plane as shown in the figure. the distance from the base, at which the particle hits the plane is close to :
(Take g = 10 ms ^–2)

The time dependence of the position of a particle of mass m = 2 is given by $$\overrightarrow r \left( t \right) = 2t\widehat i - 3{t^2}\widehat j$$ . Its angular momentum, with respect to the origin, at time t = 2 is

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is :
[Take $$\mu $$₀ = 4$$\pi $$ × 10^–7 NA^–2]