Question

A thin conducting wire of length 2m is perpendicular to x-z plane. If it is moved with velocity $\vec{v}=(2\hat{i}+3\hat{j}+4\hat{k})$ m/s through a region of magnetic induction $\vec{B}=(\hat{i}+2\hat{j}+3\hat{k})T.$ Potential difference induced between ends of the wire is

Answer 1

4V

Answer 2

The potential difference induced between the ends of a conducting wire moving in a magnetic field is given by the formula for motional electromotive force (EMF):

$\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{L}$

where:

• $\vec{v}$ is the velocity of the wire.
• $\vec{B}$ is the magnetic induction field.
• $\vec{L}$ is the vector representing the length and direction of the wire.

Given:

• Velocity, $\vec{v}=(2\hat{i}+3\hat{j}+4\hat{k})$ m/s
• Magnetic induction, $\vec{B}=(\hat{i}+2\hat{j}+3\hat{k})$ T
• Length of the wire, $L=2$ m.
• The wire is perpendicular to the x-z plane. This means the wire is oriented along the y-axis. Therefore, the length vector can be written as $\vec{L} = 2\hat{j}$ m (we can choose the positive y-direction for $\vec{L}$; the magnitude of the potential difference will be the same regardless of the direction chosen for $\vec{L}$).

Step 1: Calculate the cross product $\vec{v} \times \vec{B}$

$\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{vmatrix}$

$= \hat{i}((3)(3) - (4)(2)) - \hat{j}((2)(3) - (4)(1)) + \hat{k}((2)(2) - (3)(1))$

$= \hat{i}(9 - 8) - \hat{j}(6 - 4) + \hat{k}(4 - 3)$

$= 1\hat{i} - 2\hat{j} + 1\hat{k}$

$= \hat{i} - 2\hat{j} + \hat{k}$

Step 2: Calculate the dot product of $(\vec{v} \times \vec{B})$ with $\vec{L}$

$\varepsilon = (\hat{i} - 2\hat{j} + \hat{k}) \cdot (2\hat{j})$

$= (1)(0) + (-2)(2) + (1)(0)$

$= 0 - 4 + 0$

$= -4 \text{ V}$

The induced potential difference (EMF) is $-4$ V. The negative sign indicates the polarity of the induced EMF. The potential difference between the ends of the wire is the magnitude of this value.

Step 3: Determine the potential difference

Potential difference = $|\varepsilon| = |-4 \text{ V}| = 4 \text{ V}$.

The final answer is $\boxed{4V}$.

Explanation of the solution:

The induced potential difference (motional EMF) in a conductor moving in a magnetic field is given by $\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{L}$.

1. Identify the velocity vector $\vec{v}$, magnetic field vector $\vec{B}$, and length vector $\vec{L}$. The wire being perpendicular to the x-z plane implies $\vec{L}$ is along the y-axis, so $\vec{L} = 2\hat{j}$.

2. Calculate the cross product $\vec{v} \times \vec{B} = (2\hat{i}+3\hat{j}+4\hat{k}) \times (\hat{i}+2\hat{j}+3\hat{k}) = \hat{i} - 2\hat{j} + \hat{k}$.

3. Calculate the dot product of the result with $\vec{L}$: $\varepsilon = (\hat{i} - 2\hat{j} + \hat{k}) \cdot (2\hat{j}) = -4$ V.

4. The potential difference is the magnitude of the induced EMF, which is 4 V.