Question

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:

• A. \( -i B R \hat{j} \)
• B. \( 2 i B R \hat{j} \)
• C. \( i B R \hat{j} \)
• D. \( -2 i B R \hat{j} \)

Answer 1

The Correct Option is D

To find the magnetic force on the wire, we need to consider each part of the wire separately: the semicircular portion and the two straight sections.

1.  For the semicircular part:
   • The magnetic force on a current-carrying wire in a magnetic field is given by \(\vec{F} = i (\vec{L} \times \vec{B})\), where \(\vec{L}\) is the length vector of the wire segment.
   • For the semicircular part with radius \(R\), the length vector effectively acts along the diameter, as the summation over the circular part results in an effective length direction.
   • The effective length vector \(\vec{L}\) for the semicircle is directed downwards, and its magnitude is \(2R\).
   • The magnetic field \(\vec{B} = B_0 \hat{j}\) is perpendicular to the plane of the semicircle.
   • The force on the semicircular part is given by \(\vec{F}_{\text{semicircle}} = i (2R) B_0 \hat{k}\), as \(\vec{L} \times \vec{B}\) results in a direction out of the page, following the right-hand rule.
2. For the two straight sections:
   • Both straight segments are parallel to the magnetic field, thus the angle between them and the magnetic field is 0 degrees.
   • The force on a wire parallel to the field is zero because \(\sin(0^\circ) = 0\).
   • Therefore, \(\vec{F}_{\text{straight}} = 0\) for both straight segments.

Thus, the total force on the wire is the force due to the semicircular part alone:

\(\vec{F} = \vec{F}_{\text{semicircle}} + \vec{F}_{\text{straight}} = i (2R) B_0 \hat{k}\)

This force directed in the negative \(\hat{j}\) direction, based on the right-hand rule for a circular loop's magnetic force.

Therefore, the magnetic force on the wire is:

\(-2 i B R \hat{j}\)

The correct answer is:

\( -2 i B R \hat{j} \)

Answer 2

The wire forms a semicircular arc with radius \( R \) and length \( l = 2R \).

Given that the direction of the magnetic field \( \vec{B} \) is in the \( +\hat{k} \) direction.

The magnetic force \( \vec{F} \) is given by:

\[ \vec{F} = i \vec{l} \times \vec{B} \]

Substituting the values:

\[ \vec{F} = -2iRB \hat{j} \]