Question

A current carrying wire AC is placed in uniform transverse magnetic field then the force on wire AC :-

• A. 3N
• B. 4.2N
• C. 6N
• D. 4N

Answer 1

Answer: (A)

3N

Answer 2

The problem asks for the force on a current-carrying wire AC placed in a uniform transverse magnetic field.

Key Concept:

For a current-carrying wire of any shape placed in a uniform magnetic field, the net magnetic force on the wire depends only on the current ($I$), the magnetic field strength ($B$), and the displacement vector ($\vec{L}$) from the initial point to the final point of the wire. The formula for the force is:

$\vec{F} = I (\vec{L} \times \vec{B})$

where $\vec{L}$ is the displacement vector from point A to point C.

Step-by-step Derivations:

1. Identify the displacement vector $\vec{L}_{AC}$:

   From the given figure, the wire starts at point A and ends at point C. The vertical distance between the horizontal line passing through A and the horizontal line passing through C is 6 cm. This means the y-component of the displacement is -6 cm (since C is below A). The horizontal distance between the vertical line passing through A and the vertical line passing through C is 8 cm. This means the x-component of the displacement is +8 cm (since C is to the right of A). So, the displacement vector is $\vec{L}_{AC} = (8 \hat{i} - 6 \hat{j})$ cm.

2. Convert dimensions to SI units:

   $L_x = 8 \text{ cm} = 0.08 \text{ m}$ $L_y = -6 \text{ cm} = -0.06 \text{ m}$ So, $\vec{L}_{AC} = (0.08 \hat{i} - 0.06 \hat{j})$ m.

3. Calculate the magnitude of the effective length ($L_{eff}$):

   $L_{eff} = |\vec{L}_{AC}| = \sqrt{(0.08 \text{ m})^2 + (-0.06 \text{ m})^2}$ $L_{eff} = \sqrt{0.0064 + 0.0036}$ $L_{eff} = \sqrt{0.01}$ $L_{eff} = 0.1 \text{ m}$

4. Identify the given values:

   Current $I = 3 \text{ A}$ Magnetic field $B = 10 \text{ T}$ The magnetic field is "transverse" and indicated by $\otimes$, which means it is uniform and directed into the page. This implies the magnetic field is perpendicular to the plane containing the wire (and thus perpendicular to the displacement vector $\vec{L}_{AC}$). Therefore, the angle ($\theta$) between $\vec{L}_{AC}$ and $\vec{B}$ is $90^\circ$.

5. Calculate the magnitude of the force:

   Since $\vec{L}_{AC}$ and $\vec{B}$ are perpendicular, the magnitude of the force is given by: $F = I L_{eff} B \sin\theta$ $F = I L_{eff} B \sin(90^\circ)$ $F = I L_{eff} B$ Substitute the values: $F = (3 \text{ A}) \times (0.1 \text{ m}) \times (10 \text{ T})$ $F = 3 \times 1 \text{ N}$ $F = 3 \text{ N}$

The force on the wire AC is 3 N.

Explanation of the solution:

The force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I (\vec{L} \times \vec{B})$, where $\vec{L}$ is the displacement vector from the start to the end of the wire. From the diagram, the displacement from A to C is 8 cm horizontally and 6 cm vertically downwards. The magnitude of this effective length is $\sqrt{(8 \text{ cm})^2 + (6 \text{ cm})^2} = 10 \text{ cm} = 0.1 \text{ m}$. Since the magnetic field is uniform and perpendicular to the plane of the wire, the angle between $\vec{L}$ and $\vec{B}$ is $90^\circ$. Therefore, the magnitude of the force is $F = I L B = (3 \text{ A}) \times (0.1 \text{ m}) \times (10 \text{ T}) = 3 \text{ N}$.