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(D) Consider a small current element $dl$ on the circular loop of radius $a$. The magnetic field $B$ at this point makes an angle $	heta$ with the ve

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None

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(D) Consider a small current element $dl$ on the circular loop of radius $a$. The magnetic field $B$ at this point makes an angle $	heta$ with the vertical axis,and the current element $dl$ is perpendicular to the plane containing the radial magnetic field and the vertical axis.According to the magnetic force formula,$dF = I(dl 	imes B)$.The magnitude of the force is $dF = I dl B \sin(90^\circ) = I B dl$.By the right-hand rule,the direction of this force $dF$ is vertically upward,making an angle $	heta$ with the vertical axis,or more specifically,it acts radially outward and upward.Due to the symmetry of the loop,the horizontal components of the force on diametrically opposite elements cancel each other out.The vertical component of the force on each element $dl$ is $dF_v = dF \cos	heta = I B dl \cos	heta$.Integrating this over the entire circumference of the loop (length $2\pi a$):$F_{net} = \int dF_v = \int_0^{2\pi a} I B \cos	heta dl = I B \cos	heta \int_0^{2\pi a} dl = I B \cos	heta (2\pi a) = 2\pi aIB \cos	heta$.Since this result is not among the given options,the correct answer is None.

$A$ circular current loop of radius $a$ is placed in a radial magnetic field $B$ as shown in the figure. The net force acting on the loop is

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