A semi-circular loop of radius 30 ,cm wire ca | vedclass

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(B) The force on a current-carrying wire in a magnetic field is given by $F = I \vec{L}_{eff} 	imes \vec{B}$, where $L_{eff}$ is the effective length

Vedclass · Accepted Answer

$1.8$

Vedclass · Answer

(B) The force on a current-carrying wire in a magnetic field is given by $F = I \vec{L}_{eff} 	imes \vec{B}$, where $L_{eff}$ is the effective length (displacement vector) between the two ends of the wire.For a semi-circular loop of radius $R$, the effective length is the diameter, $L_{eff} = 2R$.Given: $R = 30 \,cm = 0.3 \,m$, $I = 6 \,A$, $B = 0.5 \,T$.The effective length $L_{eff} = 2 	imes 0.3 \,m = 0.6 \,m$.Since the magnetic field is perpendicular to the plane of the loop, the angle between the effective length vector and the magnetic field is $90^{\circ}$.Therefore, the magnitude of the force is $F = I L_{eff} B \sin(90^{\circ})$.$F = 6 \,A 	imes 0.6 \,m 	imes 0.5 \,T 	imes 1 = 1.8 \,N$.

$A$ semi-circular loop of radius $30 \,cm$ wire carries current $6 \,A$. $A$ uniform magnetic field $0.5 \,T$ is present perpendicular to the plane of the loop. What is the magnitude of force exerted on the wire (in $\,N$)?

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