Question

The magnetic moment of the arrangement shown in the figure is

• A. Zero
• B. $2\sqrt{2}M$
• C. 2 M
• D. M

Answer 1

Answer: (C)

2 M

Answer 2

The magnetic moment of a bar magnet is a vector quantity directed from its South pole to its North pole. Its magnitude is given by m * L, where m is the pole strength and L is the length of the magnet.

Let's analyze the three magnets in the arrangement:

1. Horizontal Magnet:

   • Its North pole is on the left and South pole is on the right.
   • Its length is labeled as 'M' in the figure. Let's assume its magnetic moment has a magnitude of M (the variable used in the options).
   • Since the magnetic moment vector points from S to N, for the horizontal magnet, it points to the left.
   • Let's set up a coordinate system where the bottom-left corner is the origin (0,0), the horizontal direction is the x-axis, and the vertical direction is the y-axis.
   • Magnetic moment vector of horizontal magnet: $\vec{M}_1 = -M \hat{i}$

2. Vertical Magnet:

   • Its South pole is at the bottom and North pole is at the top.
   • Its length is labeled as 'M' in the figure. So, its magnetic moment has a magnitude of M.
   • The magnetic moment vector points from S to N, so it points upwards.
   • Magnetic moment vector of vertical magnet: $\vec{M}_2 = M \hat{j}$

3. Hypotenuse Magnet:

   • Its South pole is at the bottom-left and North pole is at the top-right.
   • Its length is labeled as $\sqrt{2}M$ in the figure. If we assume the magnetic moment is proportional to length (i.e., pole strength 'm' is constant for all magnets), then its magnetic moment magnitude is $\sqrt{2}M$.
   • The triangle is a right-angled isosceles triangle (sides M and M), so the hypotenuse makes an angle of 45 degrees with the horizontal and vertical axes.
   • The magnetic moment vector points from S to N, i.e., along the hypotenuse from bottom-left to top-right.
   • We can resolve this vector into its x and y components:
     • x-component: $(\sqrt{2}M) \cos 45^\circ = (\sqrt{2}M) \frac{1}{\sqrt{2}} = M$
     • y-component: $(\sqrt{2}M) \sin 45^\circ = (\sqrt{2}M) \frac{1}{\sqrt{2}} = M$
   • Magnetic moment vector of hypotenuse magnet: $\vec{M}_3 = M \hat{i} + M \hat{j}$

Net Magnetic Moment:

The total magnetic moment of the arrangement is the vector sum of the individual magnetic moments: $\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3$ $\vec{M}_{net} = (-M \hat{i}) + (M \hat{j}) + (M \hat{i} + M \hat{j})$

Combine the $\hat{i}$ components: $(-M + M) \hat{i} = 0 \hat{i}$ Combine the $\hat{j}$ components: $(M + M) \hat{j} = 2M \hat{j}$

So, $\vec{M}_{net} = 0 \hat{i} + 2M \hat{j} = 2M \hat{j}$

The magnitude of the net magnetic moment is $|\vec{M}_{net}| = \sqrt{(0)^2 + (2M)^2} = \sqrt{4M^2} = 2M$.

The final answer is 2 M.

Explanation of the solution:

1. Identify the direction of the magnetic moment for each bar magnet (from South to North pole).
2. Assign magnitudes to each magnetic moment based on the given labels (M, M, $\sqrt{2}M$).
3. Represent each magnetic moment as a vector using a coordinate system.
   • Horizontal magnet: $\vec{M}_1 = -M \hat{i}$
   • Vertical magnet: $\vec{M}_2 = M \hat{j}$
   • Hypotenuse magnet: $\vec{M}_3 = M \hat{i} + M \hat{j}$ (resolved into components as it's at 45 degrees).
4. Calculate the net magnetic moment by vector addition: $\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3$.
5. Sum the components: $\vec{M}_{net} = (-M + M) \hat{i} + (M + M) \hat{j} = 2M \hat{j}$.
6. The magnitude of the net magnetic moment is $|2M \hat{j}| = 2M$.

The final answer is 2 M. The correct option is (3).