Question

If a straight current-carrying wire of linear density $0.12~\text{kg/m$ is suspended in mid-air by a uniform horizontal magnetic field of $0.5~\text{T}$ normal to the length of the wire, then the current through the wire is (Neglect earth’s magnetic field)}

• A. $2.4~\text{A}$
• B. $1.2~\text{A}$
• C. $0.6~\text{A}$
• D. $4.8~\text{A}$

Hint:

To balance weight using magnetic force, use $ILB = mg$ or $I = \dfrac{\lambda g}{B}$.

Answer 1

The Correct Option is A

For equilibrium: magnetic force balances weight per unit length.
$BIL = mg \Rightarrow I = \dfrac{\lambda g}{B}$
Substitute values: $\lambda = 0.12$, $g = 10$, $B = 0.5$
$I = \dfrac{0.12 \times 10}{0.5} = \dfrac{1.2}{0.5} = 2.4~\text{A}$