Question

A horizontal wire of length \(10\,\text{cm}\) and mass \(0.3\,\text{g}\) carries a current of \(5\,\text{A}\). The magnitude of the magnetic field which can keep the wire in suspension is \[ (g=10\,\text{m s}^{-2}) \]

• A. \(3\times10^{-3}\,\text{T}\)
• B. \(6\times10^{-3}\,\text{T}\)
• C. \(3\times10^{-4}\,\text{T}\)
• D. \(6\times10^{-4}\,\text{T}\)

Hint:

For a wire suspended in a magnetic field, \[ BIl=mg \] The magnetic force balances the weight of the wire.

Answer 1

The Correct Option is B

Concept: A current carrying conductor placed in a magnetic field experiences a magnetic force \[ F=BIl\sin\theta \] For suspension of the wire, \[ F=mg \] Assuming the magnetic field is perpendicular to the wire, \[ \theta=90^\circ \] and hence \[ F=BIl \]

Step 1: Write the given data. \[ m=0.3\,\text{g} = 0.3\times10^{-3}\,\text{kg} \] \[ l=10\,\text{cm} = 0.1\,\text{m} \] \[ I=5\,\text{A} \] \[ g=10\,\text{m s}^{-2} \]

Step 2: Apply the condition for suspension. \[ BIl=mg \] \[ B=\frac{mg}{Il} \] Substituting the values, \[ B = \frac{(0.3\times10^{-3})(10)} {(5)(0.1)} \] \[ B = \frac{3\times10^{-3}} {0.5} \] \[ B = 6\times10^{-3}\,\text{T} \]

Step 3: State the answer. \[ \boxed{ B=6\times10^{-3}\,\text{T} } \] Hence, the correct option is \[ \boxed{(B)} \]