A rigid wire consists of a semicircular portion of radius \( R \) and two straight sections. The wire is partially immersed in a perpendicular magnetic field \( \vec{B} = B_0 \hat{j} \) as shown in the figure.The magnetic force on the wire if it has a current \( i \) is:
To determine the magnetic force on the wire, analyze the semicircular portion and the two straight sections independently.
For the semicircular part:
The magnetic force on a current-carrying wire in a magnetic field is calculated using \(\vec{F} = i (\vec{L} \times \vec{B})\), where \(\vec{L}\) represents the length vector of the wire segment.
For the semicircular segment with radius \(R\), the effective length vector acts along the diameter due to the summation over the circular path.
The effective length vector \(\vec{L}\) for the semicircle is directed downwards and has a magnitude of \(2R\).
The magnetic field is given by \(\vec{B} = B_0 \hat{j}\) and is perpendicular to the plane of the semicircle.
The force on the semicircular part is \(\vec{F}_{\text{semicircle}} = i (2R) B_0 \hat{k}\), as the cross product \(\vec{L} \times \vec{B}\) yields a direction out of the page, consistent with the right-hand rule.
For the two straight sections:
Both straight segments are parallel to the magnetic field, resulting in an angle of 0 degrees between them and the field.
The force on a wire parallel to the magnetic field is zero because \(\sin(0^\circ) = 0\).
Consequently, \(\vec{F}_{\text{straight}} = 0\) for both straight segments.
The total force on the wire is therefore solely the force exerted on the semicircular part:
\(\vec{F} = \vec{F}_{\text{semicircle}} + \vec{F}_{\text{straight}} = i (2R) B_0 \hat{k}\)
This force is directed in the negative \(\hat{j}\) direction, as determined by the right-hand rule applied to the magnetic force of a circular loop.
The resultant magnetic force on the wire is:
\(-2 i B R \hat{j}\)
The correct answer is:
\( -2 i B R \hat{j} \)
Was this answer helpful?
0
Top Questions on Magnetic Effects of Current and Magnetism
