Question:
At a place, horizontal component of Earth's field is \( 0.3\,\text{T} \) and angle of dip is \( 60^\circ \). Total magnetic field is:
At a place, horizontal component of Earth's field is \( 0.3\,\text{T} \) and angle of dip is \( 60^\circ \). Total magnetic field is:
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Remember the components of Earth's magnetic field:
- Horizontal Component: $B_H = B \cdot \cos\theta$
- Vertical Component: $B_V = B \cdot \sin\theta$
- Ratio of components: $\tan\theta = \frac{B_V}{B_H}$
- Horizontal Component: $B_H = B \cdot \cos\theta$
- Vertical Component: $B_V = B \cdot \sin\theta$
- Ratio of components: $\tan\theta = \frac{B_V}{B_H}$
Updated On: Jun 3, 2026
- $0.3\text{ T}$
- $0.6\text{ T}$
- $0.15\text{ T}$
- $0.9\text{ T}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question asks to determine the total intensity of the Earth's magnetic field at a specific location, given its horizontal component and the angle of dip.
Step 2: Key Formula or Approach:
The total magnetic field of the Earth ($B$) is resolved into two perpendicular components: the horizontal component ($B_H$) and the vertical component ($B_V$).
The relationship between the horizontal component, the total magnetic field, and the angle of dip ($\theta$) is given by:
\[ B_H = B \cdot \cos\theta \] Rearranging this formula to solve for the total magnetic field:
\[ B = \frac{B_H}{\cos\theta} \]
Step 3: Detailed Explanation:
• The angle of dip (or magnetic inclination) is the angle made by the Earth's total magnetic field vector with the horizontal plane at that place.
• In this problem, the horizontal component of Earth's magnetic field is $B_H = 0.3\text{ T}$.
• The angle of dip is given as $\theta = 60^\circ$.
• The total magnetic field $B$ is the vector sum of its horizontal and vertical components. Since $B_H$ is the projection of $B$ along the horizontal direction, we use the cosine trigonometric ratio.
• We know that the value of $\cos(60^\circ) = 0.5 = \frac{1}{2}$.
• Substituting the values of $B_H$ and $\cos(60^\circ)$ into our rearranged equation:
\[ B = \frac{0.3}{\cos(60^\circ)} \] \[ B = \frac{0.3}{0.5} \] \[ B = 0.6\text{ T} \]
• Thus, the total magnetic field intensity at that place is $0.6\text{ T}$.
Step 4: Final Answer:
The total magnetic field of the Earth at the given place is $0.6\text{ T}$.
The question asks to determine the total intensity of the Earth's magnetic field at a specific location, given its horizontal component and the angle of dip.
Step 2: Key Formula or Approach:
The total magnetic field of the Earth ($B$) is resolved into two perpendicular components: the horizontal component ($B_H$) and the vertical component ($B_V$).
The relationship between the horizontal component, the total magnetic field, and the angle of dip ($\theta$) is given by:
\[ B_H = B \cdot \cos\theta \] Rearranging this formula to solve for the total magnetic field:
\[ B = \frac{B_H}{\cos\theta} \]
Step 3: Detailed Explanation:
• The angle of dip (or magnetic inclination) is the angle made by the Earth's total magnetic field vector with the horizontal plane at that place.
• In this problem, the horizontal component of Earth's magnetic field is $B_H = 0.3\text{ T}$.
• The angle of dip is given as $\theta = 60^\circ$.
• The total magnetic field $B$ is the vector sum of its horizontal and vertical components. Since $B_H$ is the projection of $B$ along the horizontal direction, we use the cosine trigonometric ratio.
• We know that the value of $\cos(60^\circ) = 0.5 = \frac{1}{2}$.
• Substituting the values of $B_H$ and $\cos(60^\circ)$ into our rearranged equation:
\[ B = \frac{0.3}{\cos(60^\circ)} \] \[ B = \frac{0.3}{0.5} \] \[ B = 0.6\text{ T} \]
• Thus, the total magnetic field intensity at that place is $0.6\text{ T}$.
Step 4: Final Answer:
The total magnetic field of the Earth at the given place is $0.6\text{ T}$.
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