Question

A small bar magnet lies along the x-axis with its Centre fixed at the origin. If the magnetic field at point (5\(\hat{i}\)) m due to this magnet is 4x10⁶ T,then the magnetic field at point (10\(\hat{j}\)) m will be :

• A. 2.5x 10^-7 T
• B. 2x10^-6 T
• C. 1x10^-6 T
• D. 2.0x10^-7 T
• E. 8.0x10^-8 T

Answer 1

The Correct Option is A

Given:
- The magnetic field at \((5 \hat{i})\) meters is \(4 \times 10^6 \, \text{T}\).
- We need to find the magnetic field at \((10 \hat{j})\) meters.

The magnetic field \(B\) due to a magnetic dipole at a distance \(r\) from the dipole along the axial (x-axis) direction is given by:

\[B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3}\]

And along the equatorial (y-axis) direction is given by:

\[B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3}\]

where:
- \(M\) is the magnetic moment of the bar magnet,
- \(\mu_0\) is the permeability of free space (\(\mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m}/\text{A}\)).

Given the field along the x-axis (axial position) at \(r = 5 \, \text{m}\):
\[4 \times 10^6 \, \text{T} = \frac{\mu_0}{4\pi} \frac{2M}{(5)^3}\]

Simplifying:
\[4 \times 10^6 = \frac{2 \times 10^{-7} M}{125}\]

Solving for \(M\):
\[M = \frac{4 \times 10^6 \times 125}{2 \times 10^{-7}}\]
\[M = \frac{5 \times 10^8}{10^{-7}} = 2.5 \times 10^{15} \, \text{A} \cdot \text{m}^2\]

Now, we need to find the magnetic field at \((10 \hat{j})\) meters (equatorial position):
\[B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{(10)^3}\]
\[B_{\text{equatorial}} = \frac{10^{-7} \times 2.5 \times 10^{15}}{1000}\]
\[B_{\text{equatorial}} = 2.5 \times 10^{-7} \, \text{T}\]

Therefore, the magnetic field at the point \((10 \hat{j})\) meters is \(2.5 \times 10^{-7} \, \text{T}\), 
So The correct answer is Option(A): 2.5x 10^-7 T

Answer 2

Step 1: Understand the magnetic field due to a bar magnet.

The magnetic field due to a bar magnet depends on:

• The distance \( r \) from the center of the magnet.
• The orientation of the point relative to the magnet (axial or equatorial).

For a bar magnet lying along the x-axis with its center at the origin:

• The magnetic field along the axial line is given by:

\[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3}, \]

• The magnetic field along the equatorial line is given by:

\[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3}. \] Thus, the relationship between the axial and equatorial fields at the same distance \( r \) is: \[ B_{\text{equatorial}} = \frac{B_{\text{axial}}}{2}. \] However, in this problem, the distances along the axial and equatorial lines are different, so we must account for the dependence on \( r^3 \).

 

Step 2: Analyze the given data.

We are given:

• The magnetic field at point \( (5\hat{i}) \, \text{m} \), which lies along the axial line, is \( B_{\text{axial}} = 4 \times 10^{-6} \, \text{T} \). Here, the distance is \( r_{\text{axial}} = 5 \, \text{m} \).
• We need to find the magnetic field at point \( (10\hat{j}) \, \text{m} \), which lies along the equatorial line. Here, the distance is \( r_{\text{equatorial}} = 10 \, \text{m} \).

 

Step 3: Relate the magnetic fields using the formula.

The magnetic field along the axial line is: \[ B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r_{\text{axial}}^3}. \] The magnetic field along the equatorial line is: \[ B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \cdot \frac{M}{r_{\text{equatorial}}^3}. \] Dividing these two equations, we get: \[ \frac{B_{\text{equatorial}}}{B_{\text{axial}}} = \frac{\frac{M}{r_{\text{equatorial}}^3}}{\frac{2M}{r_{\text{axial}}^3}} = \frac{r_{\text{axial}}^3}{2r_{\text{equatorial}}^3}. \] Substitute the given distances: \[ r_{\text{axial}} = 5 \, \text{m}, \quad r_{\text{equatorial}} = 10 \, \text{m}. \] Thus: \[ \frac{B_{\text{equatorial}}}{B_{\text{axial}}} = \frac{(5)^3}{2(10)^3} = \frac{125}{2(1000)} = \frac{125}{2000} = \frac{1}{16}. \] Now, solve for \( B_{\text{equatorial}} \): \[ B_{\text{equatorial}} = \frac{1}{16} \cdot B_{\text{axial}}. \] Substitute \( B_{\text{axial}} = 4 \times 10^{-6} \, \text{T} \): \[ B_{\text{equatorial}} = \frac{1}{16} \cdot 4 \times 10^{-6} = 0.25 \times 10^{-6} = 2.5 \times 10^{-7} \, \text{T}. \]

Final Answer: The magnetic field at point \( (10\hat{j}) \, \text{m} \) is \( \mathbf{2.5 \times 10^{-7} \, \text{T}} \), which corresponds to option \( \mathbf{(A)} \).