Problem 1972 - JEE Physics

Question

A magnet of magnetic moment \(50 \uparrow \mathrm{A} \mathrm{m}^{2}\) is placed along the \(\mathrm{X}\) -axis in a mag. field \(\mathrm{B}^{-}=(0.5 \uparrow+3.0 \mathrm{~J} \wedge\) ) Tesla. The torque acting on the magnet is N.m. (c) \(75 \mathrm{k} \wedge\) (d) \(25 \sqrt{5} \mathrm{k} \wedge\) (a) \(175 \mathrm{k}\) (b) \(150 \mathrm{k}\)

Step-by-Step Solution

Answer

The torque acting on the magnet is \(\boldsymbol{\tau} = 150 \hat{k}\), which corresponds to answer (b).

Step 1: Write down the given magnetic moment and magnetic field as vectors

The magnetic moment is given as \(50 \hat{i}\), which in vector form can be written as \(\boldsymbol{m} = 50 \hat{i} = (50, 0, 0)\). The magnetic field is given as \(0.5 \hat{i} + 3.0 \hat{j}\), which in vector form can be written as \(\boldsymbol{B} = 0.5 \hat{i} + 3.0 \hat{j} = (0.5, 3.0, 0)\).

Step 2: Calculate the torque by finding the cross product of the magnetic moment and magnetic field

The cross product of two vectors \((a_x, a_y, a_z)\) and \((b_x, b_y, b_z)\) is given by: \[ (a_x, a_y, a_z) \times (b_x, b_y, b_z) = (a_yb_z - a_zb_y, a_zb_x - a_xb_z, a_xb_y - a_yb_x) \] Applying this formula to the magnetic moment and magnetic field vectors, we get: \[ \boldsymbol{\tau} = (50, 0, 0) \times (0.5, 3.0, 0) = (0, 0, 50 \times 3.0 - 0 \times 0.5) = (0, 0, 150) \] So, the torque acting on the magnet is \(\boldsymbol{\tau} = 150 \hat{k} \), which corresponds to answer (b).

Key Concepts

Magnetic Moment

The magnetic moment is a fundamental concept in magnetism. It represents the strength and the orientation of a magnet's influence. Think of it as the 'arrow' that shows the direction and magnitude of a magnet's force over a distance. In physics, it's often denoted by the symbol \( \boldsymbol{m} \). This vector quantity is an essential factor in determining how a magnet will respond to external magnetic fields.

In this exercise, the magnetic moment is given as \( 50 \hat{i} \), meaning it points entirely along the x-axis. This orientation greatly influences the torque calculated later on. Understanding magnetic moments helps us predict how materials interact with magnetic fields in various applications, such as in compasses or electric motors.

A key point to remember is that the magnetic moment is all about aligning magnetic dipoles within a material, like little arrows all pointing in the same direction. Thus, being given as 50 Ampere-meter squared here, it signifies a significant amount of magnetic force pointing in a specific direction.

Cross Product

The cross product is a vector operation crucial in calculating torque, especially in electromagnetic contexts. It arises when we want to find a new vector orthogonally oriented to two given vectors. Simply put, it's a way of calculating an entirely new vector that offers an insight into rotational forces, such as torque.

Mathematically, if we have two vectors \((a_x, a_y, a_z)\) and \((b_x, b_y, b_z)\), their cross product is calculated as:

• \( (a_yb_z - a_zb_y) \)
• \( (a_zb_x - a_xb_z) \)
• \( (a_xb_y - a_yb_x) \)

The resultant vector provides a means to quantify the rotational effect one vector has on another. In the context of this exercise, applying the cross product to the magnetic moment and the magnetic field reveals the magnitude and direction of torque.

The utility of the cross product extends beyond physics into engineering, mathematics, and even computer graphics—where understanding the orientation and force in three dimensions is pivotal.

Vector Calculations

Vector calculations are indispensable in physics and engineering as they provide a framework for understanding forces, velocities, and other directional quantities. A vector is an entity with both magnitude and direction, represented by ordered components along orthogonal axes. This ordered nature makes calculations involving vectors especially powerful.

In this exercise, we see how vector calculations help simplify and solve for the torque. The magnetic moment \((50, 0, 0)\) and the magnetic field \((0.5, 3.0, 0)\) are broken down into their vector components. This breakdown allows us to apply mathematical formulas, like the cross product, systematically. Each step in the vector calculation can be traced, ensuring accuracy.

Grasping vector calculations equips you with tools to interpret complex systems, from automating machinery to understanding celestial mechanics. They form a backbone to modern computations in physics, highlighting how multi-dimensional problems can be approached logically and effectively.