Problem 1962 - JEE Physics

Question

A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. A horizontal magnetic field \(\mathrm{B}=6 \times 10^{-4}\) Tesla exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (c) \(6 \mathrm{~J}\).

Step 1: Identify the given quantities

We are given the following information: - The bar magnet's magnetic moment: \(m = 2 \times 10^4 \, \mathrm{JT^{-1}}\) - The horizontal magnetic field: \(B = 6 \times 10^{-4} \, \mathrm{T}\) (Tesla) - The angle the magnet is rotated: \(\theta = 60^{\circ}\) We need to find the work done to rotate the magnet.

Step 2: Recall the expression for work done

The expression for the work done, \(W\), in rotating a magnet in a magnetic field is given by: \[ W = -mB(\cos{\theta_1}-\cos{\theta_2})\] Where, - \(W\) is the work done, - \(m\) is the magnetic moment of the magnet, - \(B\) is the magnetic field, - \(\theta_1\) is the initial angle between the magnetic moment and the magnetic field, - \(\theta_2\) is the final angle between the magnetic moment and the magnetic field. In this problem, the magnet is initially parallel to the field (\(\theta_1 = 0^{\circ}\)) and is finally at an angle of \(60^{\circ}\) (\(\theta_2 = 60^{\circ}\)) from the field.

Step 3: Calculate the work done

Now, we can plug in the given values into the formula for the work done: \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(\cos{0^{\circ}} - \cos{60^{\circ}})\] To calculate the value, we must find the cosine of the angles. Remember that \(\cos{0^{\circ}} =1\) and \(\cos{60^{\circ}} = \frac{1}{2}\). \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(1 - \frac{1}{2})\] Now, we can perform the calculations: \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(\frac{1}{2})\] \[W = -\frac{1}{2}(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})\] \[W = -\frac{1}{2}(12\,\mathrm{J})\] \[W = -6\,\mathrm{J}\] Here, the negative sign indicates that work has been done on the magnet to rotate it. The absolute value of the work done is \(6 \, \mathrm{J}\).

Step 4: Match the result with the options

Compare the calculated work done to the given options: (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\) Option (c) matches our calculated work done of \(6 \, \mathrm{J}\). Therefore, the correct answer is (c) \(6 \mathrm{~J}\).

Key Concepts

Magnetic Moment

Every bar magnet, like the one in our exercise, possesses a magnetic moment. Think of it as the strength and orientation of the magnet's internal magnetism. It is a vector quantity with both a magnitude and direction.

• The magnitude indicates how strong the magnetic field is that the magnet can produce.
• The direction points from the magnet's south pole to its north pole.

The magnetic moment, often designated by the symbol \( m \), can be expressed in units of joules per tesla (JT^-1). Here, the bar magnet's magnetic moment is given as \(2 \times 10^4\, \mathrm{JT^{-1}} \). This high value means the magnet is quite strong. Understanding the magnetic moment is crucial because it interacts with external magnetic fields, influencing how much work is done, as we explore in this specific problem.

Magnetic Field

A magnetic field is an invisible force field that surrounds magnetized objects and electric currents. It is represented by the symbol \( B \) and is measured in units called Tesla (T).

• This force field exerts a magnetic influence on other magnetized objects and moving charges within it.
• The strength and direction of a magnetic field can vary depending on the source and other influencing factors.

In our exercise, a horizontal magnetic field of \(6 \times 10^{-4}\, \mathrm{T}\) is present. This means that in any point around the space described, the magnetic field has a small, yet measurable influence. When the bar magnet is placed in this field, it experiences a torque, causing it to align or rotate, based on its magnetic moment. This interaction is at the heart of calculating the work done during the rotation.

Rotation of Magnet

Rotating a magnet in a magnetic field involves some remarkable physics. When you place a magnet in a magnetic field and then turn it from one angle to another, like from parallel to 60 degrees, you do work. This concept arises from the alignment processes and interactions between the magnetic moment and the field.

• The initial alignment is where the magnet is parallel to the magnetic field's direction, often a state of minimal energy.
• As it rotates to a new angle, external work is needed to overcome the resistance from the magnetic forces trying to keep it aligned.

The work done in this context can be quantified using the equation: \[W = -mB(\cos{\theta_1} - \cos{\theta_2})\] This formula shows that work is related to the change in cosine values of the angles concerning the initial and final rotations. For our magnet, going from \(0^{\circ}\) to \(60^{\circ}\) involves overcoming these resistances, and the calculations show an absolute work amount of 6 Joules. This process of rotation and the accompanying work done demonstrate the dynamic interaction between the magnetic moment and external fields.