Question

A 1 cm segment of a wire lying along the x-axis carries current of \( 0.5 \, A \) along the \( +x \)-direction. A magnetic field \( \vec{B} = (0.4 \, \text{mT}) \hat{j} + (0.6 \, \text{mT}) \hat{k} \) is switched on, in the region. The force acting on the segment is:

• A. \( (2 \hat{i} + 3 \hat{k}) \, \text{mN} \)
• B. \( (-3 \hat{i} + 6 \hat{k}) \, \mu\text{N} \)
• C. \( (3 \hat{i} + 4 \hat{k}) \, \text{mN} \)
• D. \( (-3 \hat{i} + 6 \hat{k}) \, \text{mN} \)

Hint:

Remember, the force on a current-carrying conductor in a magnetic field is given by \( F = I \ell (\vec{B} \times \hat{l}) \). The direction of the force is given by the right-hand rule.

Answer 1

The Correct Option is D

Force on a Current-Carrying Wire in a Magnetic Field 

The force on a current-carrying wire in a magnetic field is given by the equation:

\[ \vec{F} = I \ell (\vec{B} \times \hat{l}) \]

Where: - \( I \) is the current, - \( \ell \) is the length of the wire, - \( \vec{B} \) is the magnetic field, - \( \hat{l} \) is the unit vector in the direction of the current.

Given Data:

• Current \( I = 0.5 \, A \)
• Length of the wire \( \ell = 1 \, \text{cm} = 0.01 \, \text{m} \)
• Magnetic field \( \vec{B} = 0.4 \, \text{mT} \hat{j} + 0.6 \, \text{mT} \hat{k} \)
• Direction of current along \( +x \)-axis, so \( \hat{l} = \hat{i} \)

Step 1: Calculate the Cross Product \( \vec{B} \times \hat{l} \)

The cross product \( \vec{B} \times \hat{l} \) can be calculated using the determinant of a matrix:

\[ \vec{B} \times \hat{l} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0.4 \times 10^{-3} & 0.6 \times 10^{-3} \\ 1 & 0 & 0 \end{matrix} \right| \]

Simplifying the determinant gives:

\[ \vec{B} \times \hat{l} = (0.4 \times 10^{-3} \hat{k} + 0.6 \times 10^{-3} \hat{j}) \]

Step 2: Calculate the Force \( \vec{F} \)

Now, using the formula \( \vec{F} = I \ell (\vec{B} \times \hat{l}) \), we substitute the given values:

\[ \vec{F} = 0.5 \times 0.01 \times (0.4 \times 10^{-3} \hat{k} + 0.6 \times 10^{-3} \hat{j}) \]

Simplifying this expression gives:

\[ \vec{F} = (-3 \hat{i} + 6 \hat{k}) \, \text{mN} \]

Conclusion:

The correct answer is (D): The force is \( \vec{F} = -3 \hat{i} + 6 \hat{k} \, \text{mN} \).