The figure shows a conducting loop ADCA carry | vedclass

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(D) The magnetic force on any current-carrying wire segment in a uniform magnetic field is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{

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$2 i R B_0$

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(D) The magnetic force on any current-carrying wire segment in a uniform magnetic field is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the vector displacement from the start point to the end point of the segment.For the semicircular part $ADC$,the starting point is $A$ and the ending point is $C$.The straight-line distance (displacement) between $A$ and $C$ is the diameter of the semicircle,which is $L = 2R$.Since the magnetic field $B_0$ is uniform and perpendicular to the plane of the loop,the magnitude of the force is $F = iLB_0$.Substituting $L = 2R$,we get $F = i(2R)B_0 = 2iRB_0$.Thus,the magnitude of the force on the semicircular part is $2iRB_0$.

The figure shows a conducting loop $ADCA$ carrying current $i$ and placed in a region of uniform magnetic field $B_0$. The part $ADC$ forms a semicircle of radius $R$. The magnitude of the force on the semicircular part of the loop is equal to

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