Question

A straight wire of mass \(250\,\text{g}\) and length \(2.5\,\text{m}\) carries a current of \(4\,\text{A}\). It can be suspended in mid air by a uniform horizontal magnetic field of magnitude:

• A. \(0.145\,\text{T}\)
• B. \(0.245\,\text{T}\)
• C. \(0.625\,\text{T}\)
• D. \(2.245\,\text{T}\)

Hint:

For a current-carrying conductor in magnetic field, maximum force occurs when wire is perpendicular to field: \(F = BIL\).

Answer 1

The Correct Option is B

Step 1: Understand the condition for suspension.
For the wire to be suspended in air, the magnetic force must balance the weight of the wire.
\[ F_{\text{magnetic}} = mg \]

Step 2: Use magnetic force formula.
\[ F = BIL \]
So,
\[ BIL = mg \]

Step 3: Convert given values into SI units.
\[ m = 250\,\text{g} = 0.25\,\text{kg} \]
\[ I = 4\,\text{A}, \quad L = 2.5\,\text{m}, \quad g = 9.8\,\text{m/s}^2 \]

Step 4: Substitute values.
\[ B \times 4 \times 2.5 = 0.25 \times 9.8 \]

Step 5: Simplify.
\[ 10B = 2.45 \]
\[ B = \frac{2.45}{10} \]
\[ B = 0.245\,\text{T} \]

Step 6: Interpretation.
The magnetic field provides an upward force balancing the downward gravitational force.

Step 7: Final answer.
\[ \boxed{0.245\,\text{T}} \]