Question

A straight wire of mass \(200\,\text{g}\) and length \(1.5\,\text{m}\) carries a current of \(2\,\text{A}\). It is suspended in mid-air by a uniform horizontal magnetic field \(B\). The magnitude of \(B\) is \((g=10\,\text{m s}^{-2})\):

• A. \(2\,\text{T}\)
• B. \(1.5\,\text{T}\)
• C. \(0.55\,\text{T}\)
• D. \(0.67\,\text{T}\)

Hint:

For a current-carrying wire suspended by a magnetic field: \[ BIL=mg \] Hence, \[ \boxed{B=\frac{mg}{IL}} \] This formula is frequently asked in CUET, JEE Main and NEET Physics examinations.

Answer 1

The Correct Option is D

Step 1: Apply the condition for suspension. For the wire to remain suspended in air, the magnetic force must balance its weight. \[ F_B=mg \]

Step 2: Use magnetic force on a current-carrying conductor. The magnetic force is given by: \[ F_B=BIL\sin\theta \] For maximum upward force, \[ \theta=90^\circ \] Therefore, \[ F_B=BIL \]

Step 3: Substitute the given values. Mass: \[ m=200\,\text{g}=0.2\,\text{kg} \] Weight: \[ mg=0.2\times10=2\,\text{N} \] Using \[ BIL=mg \] \[ B\times2\times1.5=2 \] \[ 3B=2 \] \[ B=\frac{2}{3}\,\text{T} \] \[ B=0.67\,\text{T} \] Therefore, \[ \boxed{B=0.67\,\text{T}} \] Hence, the correct answer is: \[ \boxed{\text{(D)}} \]