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Question: A thin wire of length 2m is perpendicular to the xy plane. It is moved with velocity $\overrightarro...

A thin wire of length 2m is perpendicular to the xy plane. It is moved with velocity v→=(2i^+3j^+k^)\overrightarrow{v}=(2\hat{i}+3\hat{j}+\hat{k}) m/s through a region of magnetic induction B→=(i^+2j^)\overrightarrow{B}=(\hat{i}+2\hat{j}) Wb/m2^2. Then potential difference induced between the ends of the wire :

A

2 volts

B

4 volts

C

0 volts

D

none of these

Answer

2 volts

Explanation

Solution

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):

ε=(v⃗×B⃗)⋅L⃗\varepsilon = (\vec{v} \times \vec{B}) \cdot \vec{L}

where:

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

Given:

  • Velocity, v⃗=(2i^+3j^+k^)\vec{v}=(2\hat{i}+3\hat{j}+\hat{k}) m/s
  • Magnetic induction, B⃗=(i^+2j^)\vec{B}=(\hat{i}+2\hat{j}) Wb/m2^2
  • Length of the wire, L=2L=2 m.
  • The wire is perpendicular to the xy plane. This means the wire is oriented along the z-axis. Therefore, the length vector can be written as L⃗=2k^\vec{L} = 2\hat{k} m (we can choose the positive z-direction for L⃗\vec{L}; the magnitude of the potential difference will be the same regardless of the direction chosen for L⃗\vec{L}).

Step 1: Calculate the cross product v⃗×B⃗\vec{v} \times \vec{B}

v⃗×B⃗=∣i^j^k^231120∣\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 0 \end{vmatrix}

=i^((3)(0)−(1)(2))−j^((2)(0)−(1)(1))+k^((2)(2)−(3)(1))= \hat{i}((3)(0) - (1)(2)) - \hat{j}((2)(0) - (1)(1)) + \hat{k}((2)(2) - (3)(1))

=i^(0−2)−j^(0−1)+k^(4−3)= \hat{i}(0 - 2) - \hat{j}(0 - 1) + \hat{k}(4 - 3)

=−2i^+j^+k^= -2\hat{i} + \hat{j} + \hat{k}

Step 2: Calculate the dot product of (v⃗×B⃗)(\vec{v} \times \vec{B}) with L⃗\vec{L}

ε=(−2i^+j^+k^)⋅(2k^)\varepsilon = (-2\hat{i} + \hat{j} + \hat{k}) \cdot (2\hat{k})

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

=0+0+2= 0 + 0 + 2

=2 V= 2 \text{ V}

The induced potential difference (EMF) is 22 V. The potential difference between the ends of the wire is the magnitude of this value.

Step 3: Determine the potential difference

Potential difference = ∣ε∣=∣2 V∣=2 V|\varepsilon| = |2 \text{ V}| = 2 \text{ V}.