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(A) The magnetic force on a current-carrying wire segment is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the displacement vector

Vedclass · Accepted Answer

No resultant force acts on the loop

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(A) The magnetic force on a current-carrying wire segment is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the displacement vector from the start to the end of the segment.For the segment $PQR$,the starting point is $P$ and the ending point is $R$. Thus,the effective length vector is $\vec{L} = \vec{PR}$.The magnitude of the force is $F = I L_{PR} B \sin(	heta)$,where $	heta$ is the angle between $\vec{PR}$ and $\vec{B}$.From the geometry,$PR = PQ 	an(60^\circ) = PQ \sqrt{3}$.The magnetic field $\vec{B}$ is directed to the right. The vector $\vec{PR}$ is horizontal (to the right). Therefore,the angle between $\vec{PR}$ and $\vec{B}$ is $0^\circ$.Since $\sin(0^\circ) = 0$,the force on the segment $PR$ is $0$.However,the question asks for the force on the segment $PQR$. The force on $PQR$ is the vector sum of forces on $PQ$ and $QR$.Force on $PQ$: $\vec{F}_{PQ} = I(\vec{PQ} 	imes \vec{B})$. $\vec{PQ}$ is upward,$\vec{B}$ is to the right. By the right-hand rule,$\vec{PQ} 	imes \vec{B}$ is into the page. Magnitude $F_{PQ} = I(PQ)B \sin(90^\circ) = I(PQ)B$.Force on $QR$: $\vec{F}_{QR} = I(\vec{QR} 	imes \vec{B})$. The angle between $\vec{QR}$ and $\vec{B}$ is $150^\circ$ (since the angle at $Q$ is $60^\circ$). Magnitude $F_{QR} = I(QR)B \sin(150^\circ) = I(QR)B(1/2)$. Direction is into the page.Total force $F = F_{PQ} + F_{QR} = I(PQ)B + I(QR)B/2$. Given $QR = PQ / \cos(60^\circ) = 2PQ$. So $F = I(PQ)B + I(2PQ)B/2 = 2I(PQ)B$. This does not match options directly. Re-evaluating: The force on any arbitrary wire segment is $I(\vec{L}_{eff} 	imes \vec{B})$. For $PQR$,$\vec{L}_{eff} = \vec{PR}$. Since $\vec{PR}$ is parallel to $\vec{B}$,the force is $0$.

For the circuit shown in the figure,the direction and magnitude of the force on the segment $PQR$ is:

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