/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 A short wire \(A B\) carrying \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 accelerations at points A and B on the wire are the same and non-zero, i.e., 'a_{A} = a_{B} ≠ 0'. So, the correct option is (C).

Step by step solution

01

Recall the principle of magnetic interaction

Whenever a wire carrying current is placed inside a magnetic field, it experiences a magnetic force. This force depends on the strength of the magnetic field which in turn depends on the distance from the long wire producing the field.
02

Calculate the force on the wire AB

Every section of the wire AB will experience the same force per unit length, since the wire is perpendicular to the long wire and thus equally distant from it. Though the total force is maximized at the ends and zero in the middle, the wire as a whole experiences a uniform field and thus the same force throughout.
03

Relate Force and Acceleration

Since Force equals mass times acceleration (F = m*a), the uniform force experienced by the wire will result in an uniform acceleration throughout the wire. This means the acceleration at points A and B will be the same.
04

Find the Answer

The accelerations at A and B are equal and non-zero, assuming there is current flowing in both the wires. Hence, 'a_{A} = a_{B} ≠ 0'. This corresponds to option (C).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Force
The concept of magnetic force is fundamental to understanding how current-carrying wires interact. When a wire carrying an electric current is placed in a magnetic field, it experiences a force perpendicular to both the direction of the current and the magnetic field. This is due to the Lorentz force law, where the magnetic force on a segment of wire is given by the equation:\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where:
  • \( I \) is the current flowing through the wire,
  • \( L \) is the length of the wire within the magnetic field,
  • \( B \) is the magnetic field strength, and
  • \( \theta \) is the angle between the wire and the magnetic field.
In our scenario, since wire AB is perpendicular to the magnetic field created by the long vertical wire, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \). This simplifies our calculations, making the magnetic force on the wire AB directly proportional to the current and the length of the wire.
Acceleration and Force
In physics, understanding the relationship between force and acceleration is crucial, particularly under Newton's Second Law of Motion, which states that force equals mass times acceleration (\( F = m \cdot a \)). This relationship shows that an object will accelerate proportionally to the force applied to it, and inversely with its mass.

For wire AB in this exercise, each point along the wire experiences the same magnetic force because the wire lies uniformly in the magnetic field generated by the long wire. Due to this uniform force, the acceleration at both ends of the wire, points A and B, is the same, as seen from:\[ a_{A} = a_{B} = \frac{F}{m} \]This equality holds because we assume both ends of the wire have the same mass and experience the uniform force. Hence, in the absence of other forces, such as gravity or additional external forces, both ends of the wire will accelerate equally.
Current-Carrying Wire
A current-carrying wire, such as AB in our exercise, behaves in an interesting manner when interacting with a magnetic field. The wire itself becomes a source of its own magnetic field, interacting with existing magnetic fields. The interaction between these fields results in a force on the wire, as described by the magnetic force principle.
  • Current in the wire generates a circular magnetic field around it.
  • When placed near another wire, this field interacts, creating forces according to their relative current directions.
  • The direction of the force experienced by a section of wire can be determined by the "right-hand rule," which concludes that if your thumb points in the direction of the current and your fingers point in the direction of the magnetic field, the palm will face the direction of the force exerted on the wire.
Since wire AB is in a gravity-free space, it can freely accelerate due to the magnetic forces without interference, demonstrating the elegance of magnetic interactions between two current-carrying wires.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A conducting \(\operatorname{rod} A B\) of length \(l=1 \mathrm{~m}\) is moving at a velocity \(v=4 \mathrm{~m} / \mathrm{s}\) making an angle \(30^{\circ}\) with its length. \(C\) is the middle point of the rod. A uniform magnetic field \(B=2 \mathrm{~T}\) exists in a direction perpendicular to the plane of motion, then (A) \(v_{A}-v_{B}=8 \mathrm{v}\) (B) \(v_{A}-v_{B}=4 \mathrm{v}\) (C) \(v_{B}-v_{C}=2 \mathrm{v}\) (D) \(v_{B}-v_{C}=-2 \mathrm{v}\)

A magnet makes 30 oscillations per minute at a plane where intensity is \(32 \mathrm{~T}\). At another place it takes \(1 \mathrm{~s}\) to complete one oscillation. The value of horizontal intensity at the second place is, (A) \(12.8 \mathrm{~T}\) (B) \(25.6 \mathrm{~T}\) (C) \(128 \mathrm{~T}\) (D) \(256 \mathrm{~T}\)

Curie temperature is the temperature above which [2003] (A) a ferromagnetic material becomes paramagnetic. (B) a paramagnetic material becomes diamagnetic. (C) a ferromagnetic material becomes diamagnetic. (D) a paramagnetic material becomes ferromagnetic.

There is a small metallic ring of radius \(l_{0}\) having negligible resistance placed perpendicular to a constant magnetic field \(B_{0}\). One end of a rod is hinged at the centre of ring \(O\) and other end is placed on the ring. Now rod is rotated with constant angular velocity \(\omega_{0}\) by some external agent and circuit is connected as shown in Fig. \(15.55\), initially switch is open and capacitor is uncharged. If switch \(S\) is closed at \(t=0\), then calculate heat loss from the resistor \(R_{2}\) from \(t=0\) to the instant when voltage across the capacitor becomes \(v_{0}\) (Assume plane of ring to be horizontal and friction to be an absent at all the contacts). (Assume, \(R_{2}=2 R_{1}\), \(\left.B_{0} l_{0}^{2} \omega_{0}=4 v_{0}\right)\) (A) \(\frac{1}{2} C v_{0}^{2}\) (B) \(\frac{1}{6} C v_{0}^{2}\) (C) \(\frac{2}{3} C v_{0}^{2}\) (D) \(\frac{1}{3} C v_{0}^{2}\)

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.