Question

A straight wire of mass \(500\,\text{g}\) and length \(7\,\text{m}\) carries a current of \(1.4\,\text{A}\). It is suspended in mid-air by a uniform horizontal magnetic field \(B\) perpendicular to the length of the wire. The magnitude of the magnetic field is:

• A. \(3.5\,\text{T}\)
• B. \(0.7\,\text{T}\)
• C. \(0.5\,\text{T}\)
• D. \(2.5\,\text{T}\)

Hint:

For a wire suspended by magnetic force: \[ BIL=mg \] Hence, \[ \boxed{B=\frac{mg}{IL}} \] When the magnetic field is perpendicular to the wire, \[ \sin 90^\circ =1 \] which gives the maximum magnetic force.

Answer 1

The Correct Option is C

Step 1: Apply the condition for suspension. Since the wire is suspended in air, the magnetic force balances its weight. \[ F_B=mg \]

Step 2: Use the magnetic force formula. Magnetic force on a current-carrying conductor is: \[ F_B=BIL\sin\theta \] Since the magnetic field is perpendicular to the wire, \[ \theta=90^\circ \] and therefore, \[ F_B=BIL \]

Step 3: Substitute the given values. \[ m=500\,\text{g}=0.5\,\text{kg} \] Taking \[ g=9.8\,\text{m s}^{-2} \] \[ mg=0.5\times9.8=4.9\,\text{N} \] Using \[ BIL=mg \] \[ B\times1.4\times7=4.9 \] \[ 9.8B=4.9 \] \[ B=\frac{4.9}{9.8} \] \[ B=0.5\,\text{T} \] Therefore, \[ \boxed{B=0.5\,\text{T}} \] Hence, the correct answer is: \[ \boxed{\text{(C)}} \]