The horizontal component of earth’s magnetic field at a place is \( 3.5 \times 10^{-5} \, \text{T} \). A very long straight conductor carrying current of \( \sqrt{2} \, \text{A} \) in the direction from South East to North West is placed. The force per unit length experienced by the conductor is \( \ldots \times 10^{-6} \, \text{N/m} \).
Correct Answer: 35
Approach Solution - 1
The problem asks for the magnetic force per unit length experienced by a long, straight current-carrying conductor placed in the Earth's horizontal magnetic field.
Concept Used:
The magnetic force (\( \vec{F} \)) on a straight conductor of length \(L\) carrying a current \(I\) in a uniform magnetic field \( \vec{B} \) is given by the Lorentz force formula:
\[ \vec{F} = I (\vec{L} \times \vec{B}) \]The magnitude of this force is given by:
\[ F = I L B \sin(\theta) \]where \( \theta \) is the angle between the direction of the current (vector \( \vec{L} \)) and the direction of the magnetic field (\( \vec{B} \)).
The force per unit length (\( \frac{F}{L} \)) is therefore:
\[ \frac{F}{L} = I B \sin(\theta) \]Step-by-Step Solution:
Step 1: Identify the given physical quantities.
- Magnitude of the horizontal component of Earth's magnetic field, \( B_H = 3.5 \times 10^{-5} \, \text{T} \).
- Current in the conductor, \( I = \sqrt{2} \, \text{A} \).
Step 2: Determine the directions and the angle between the current and the magnetic field.
The direction of the horizontal component of Earth's magnetic field (\( \vec{B}_H \)) is from geographic South to geographic North.
The direction of the current (\( I \)) is from South-East (SE) to North-West (NW).
Let's visualize these directions on a compass:
- The North direction is at \( 0^\circ \).
- The North-West direction is at an angle of \( 45^\circ \) to the West of North.
Therefore, the angle (\( \theta \)) between the direction of the magnetic field (North) and the direction of the current (North-West) is \( 45^\circ \).
Step 3: Substitute the known values into the formula for force per unit length.
\[ \frac{F}{L} = I B_H \sin(\theta) \] \[ \frac{F}{L} = (\sqrt{2} \, \text{A}) \times (3.5 \times 10^{-5} \, \text{T}) \times \sin(45^\circ) \]Final Computation & Result:
Step 4: Calculate the numerical value.
We know that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \).
\[ \frac{F}{L} = \sqrt{2} \times 3.5 \times 10^{-5} \times \frac{1}{\sqrt{2}} \, \text{N/m} \]The \( \sqrt{2} \) terms cancel out.
\[ \frac{F}{L} = 3.5 \times 10^{-5} \, \text{N/m} \]The problem asks for the answer to be expressed in the form \( \ldots \times 10^{-6} \, \text{N/m} \).
\[ 3.5 \times 10^{-5} = 35 \times 10^{-6} \]The force per unit length experienced by the conductor is 35 \( \times 10^{-6} \, \text{N/m} \).
Approach Solution -2
Calculate Magnetic Force per Unit Length: The force per unit length \( \frac{F}{\ell} \) on a current-carrying conductor in a magnetic field is given by:
\[ \frac{F}{\ell} = i B \sin \theta \]
where:
\( i = \sqrt{2} \, \text{A} \) (current in the conductor)
\( B = 3.5 \times 10^{-5} \, \text{T} \) (magnetic field)
\( \theta = 45^\circ \) (angle between current direction and magnetic field)
Substitute Values: Using \( \sin 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \frac{F}{\ell} = (\sqrt{2}) \times (3.5 \times 10^{-5}) \times \frac{1}{\sqrt{2}} = 35 \times 10^{-6} \, \text{N/m} \]
Conclusion: The force per unit length experienced by the conductor is:
\[ 35 \times 10^{-6} \, \text{N/m} \]
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