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Chapter 15: Problem 89

A short wire \(A B\) carrying \(I_{1}\) current lies in the plane of long wire which carry \(I\) current upward. If wire \(A B\) is released from horizontal position and \(a_{A}\) and \(a_{B}\) are magnitude of acceleration of points \(A\) and \(B\), respectively, then select the correct alternative. (The space is gravity-free.) (A) \(a_{A}>a_{B}\) (B) \(a_{A}

Short Answer

Expert verified
The correct answer is (C) \(a_{A} = a_{B} \neq 0\). This is because both points A and B on the wire will experience the same acceleration as a result of the even distribution of the electromagnetic force, and the acceleration is not zero due to the external current.

Step by step solution

01

Consider wire AB in the presence of magnetic field

Wire AB is exposed to a magnetic field produced by the long wire carrying an \(I\) current. Since the currents \(I_{1}\) and \(I\) have parallel directions (both are moving upward), the magnetic field at point A and B will result in a normal force on wire AB towards the long wire in their respective directions.
02

Analyze the impact on points A and B

Since there is no gravity and the magnetic field acts perpendicular to the wire AB, the entire wire AB experiences force in the same direction and to the same degree. Therefore, accelerations at both points A and B will be the same due to the homogeneous force distribution.
03

Choose the correct alternative

Since points A and B experience the same acceleration due to identical forces, and the acceleration is non-zero because there is an external current inducing magnetic field, the correct option is (C) \(a_{A} = a_{B} \neq 0\).

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Most popular questions from this chapter

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