Question

A wire carrying current \(i\), bent as shown in the figure, is placed in a uniform magnetic field \(B\) directed normally out of the plane of the figure. The force on the wire is:

• A. \(4BiR\), directed vertically downward
• B. \(3BiR\), directed vertically upward
• C. \(Bi(2R+\pi R)\), directed vertically downward
• D. \(2BiR\), from \(P\) to \(Q\)

Hint:

For any shaped wire placed in a uniform magnetic field, \[ \boxed{\vec F=I\,(\vec r_{\text{final}}-\vec r_{\text{initial}})\times\vec B} \] The force depends only on the displacement between the endpoints, not on the actual shape of the wire. This is a very common CUET/JEE conceptual question.

Answer 1

The Correct Option is A

Step 1: Use the force formula for a wire in a uniform magnetic field. For a wire carrying current in a uniform magnetic field, \[ \vec F = I\int d\vec l \times \vec B \] Since \(\vec B\) is uniform, \[ \vec F = I(\vec L)\times\vec B \] where \(\vec L\) is the displacement vector joining the initial and final points of the wire.

Step 2: Find the effective displacement. The wire consists of:
• Left straight segment of length \(R\)
• Semicircular arc of radius \(R\)
• Right straight segment of length \(R\) The horizontal displacement between the two end points is \[ R + 2R + R = 4R \] Thus, \[ L=4R \] directed horizontally.

Step 3: Calculate the magnitude of force. Since \(\vec L\) lies in the plane of the paper and \(\vec B\) is perpendicular to the plane, \[ \theta=90^\circ \] Therefore, \[ F=BIL\sin 90^\circ \] \[ F=BI(4R) \] \[ F=4BiR \]

Step 4: Determine the direction. Current flows from left to right and magnetic field is out of the plane. Using the right-hand rule: \[ \vec F=\vec L\times\vec B \] the force acts vertically downward. Therefore, \[ \boxed{F=4BiR} \] directed vertically downward. Hence, the correct answer is: \[ \boxed{\text{(A)}} \]