Question

A 1 cm segment of a wire lying along the x-axis carries a 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. The force acting on the segment is:

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

Hint:

The force on a current-carrying conductor in a magnetic field is given by \( \vec{F} = I L \vec{B} \times \hat{l} \), where \( \hat{l} \) is the unit vector along the wire direction.

Answer 1

The Correct Option is B

The magnetic force on a current-carrying wire in a magnetic field is calculated using the formula \(\vec{F} = I (\vec{L} \times \vec{B})\).

Given values are:

• Current, \(I = 0.5 \, \text{A}\)
• Wire length, \(L = 1 \, \text{cm}\) which is \(0.01 \, \text{m}\)
• Direction of the wire is along the \(+x\)-axis, hence \(\vec{L} = 0.01 \, \hat{i}\, \text{m}\)
• Magnetic field, \(\vec{B} = (0.4 \, \text{mT} \hat{j}) + (0.6 \, \text{mT} \hat{k})\)
• Conversion of millitesla to tesla: \(1 \, \text{mT} = 10^{-3} \, \text{T}\)
• Therefore, \(\vec{B} = (0.4 \times 10^{-3} \hat{j} + 0.6 \times 10^{-3} \hat{k}) \, \text{T}\)

The cross product \(\vec{L} \times \vec{B}\) is computed as follows:

\(\vec{L} \times \vec{B} = (0.01 \, \hat{i}) \times [(0.4 \times 10^{-3} \hat{j}) + (0.6 \times 10^{-3} \hat{k})]\)

Utilizing the vector cross product rules:

\((\hat{i} \times \hat{j}) = \hat{k}\)

\((\hat{i} \times \hat{k}) = -\hat{j}\)

This results in:

\(\vec{L} \times \vec{B} = 0.01 \{(0.4 \times 10^{-3}) \hat{k} - (0.6 \times 10^{-3}) \hat{j}\}\)

\(\vec{L} \times \vec{B} = (0.4 \times 10^{-5} \hat{k} - 0.6 \times 10^{-5} \hat{j})\)

Now, the force \(\vec{F} = I (\vec{L} \times \vec{B})\) is calculated:

\(\vec{F} = 0.5 (0.4 \times 10^{-5} \hat{k} - 0.6 \times 10^{-5} \hat{j})\)

\(\vec{F} = (0.2 \times 10^{-5} \hat{k} - 0.3 \times 10^{-5} \hat{j})\)

This can be written as:

\(\vec{F} = (2 \hat{k} - 3 \hat{j}) \times 10^{-6} \text{N}\)

Expressing the force in micronewtons (\(\mu\text{N}\)):

\(\vec{F} = (-3 \hat{j} + 2 \hat{k}) \, \mu\text{N}\)

The resultant force acting on the wire segment is \( (-3 \hat{j} + 2 \hat{k}) \, \mu\text{N} \).