Question

If a straight wire of length \(175\,\text{cm}\) and mass \(250\,\text{g}\) is suspended freely in a uniform horizontal magnetic field of \(0.7\,\text{T}\), then the current flowing through the wire is:

• A. \(20\,\text{A}\)
• B. \(0.02\,\text{A}\)
• C. \(2\,\text{A}\)
• D. \(0.2\,\text{A}\)

Hint:

For a wire suspended by magnetic force: \[ BIL=mg \] Hence, \[ \boxed{I=\frac{mg}{BL}} \] Always convert: \[ \text{cm} \rightarrow \text{m} \] and \[ \text{g} \rightarrow \text{kg} \] before substitution.

Answer 1

The Correct Option is C

Step 1: Apply the condition for suspension. For the wire to remain suspended, \[ F_B=mg \]

Step 2: Use the magnetic force formula. Magnetic force on a current-carrying conductor is: \[ F_B=BIL\sin\theta \] For maximum upward force, \[ \theta=90^\circ \] Thus, \[ BIL=mg \]

Step 3: Substitute the given values. \[ m=250\,\text{g}=0.25\,\text{kg} \] \[ L=175\,\text{cm}=1.75\,\text{m} \] \[ B=0.7\,\text{T} \] Taking \[ g=9.8\,\text{m s}^{-2} \] \[ I=\frac{mg}{BL} \] \[ I = \frac{0.25\times9.8} {0.7\times1.75} \] \[ I = \frac{2.45}{1.225} \] \[ I=2\,\text{A} \] Therefore, \[ \boxed{I=2\,\text{A}} \] Hence, the correct answer is: \[ \boxed{\text{(C)}} \]