Wire Equilibrium in Magnetic Field – Detailed Solution

Wire in Uniform Magnetic Field

A straight wire \(AB\) of mass \(100\,\text{g}\) (0.1 kg) and length \(100\,\text{cm}\) (1.0 m) is suspended by two flexible leads. It is placed in a uniform magnetic field of magnitude \(0.2\,\text{T}\), directed into the page, as shown in the figure.

Question: A Straight wire AB of mass 100gm and length 100 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.2 T as shown in fig. the magnitude of the current required in the wire to remove tension in the supporting leads (i.e., to balance the wire's weight). Take \(g = 10\,\text{m/s}^2\).

Possible Answers (in A):

(1) 2 A

(2) 3 A

(3) 5 A

(4) 10 A

Step-by-Step Explanation

1. Given Data:
   - Mass of wire, \(m = 0.1\,\text{kg}\)
   - Length of wire, \(\ell = 1.0\,\text{m}\)
   - Magnetic field, \(B = 0.2\,\text{T}\)
   - \(g = 10\,\text{m/s}^2\)
   - Let the required current be \(I\).
2. For Equilibrium:
   The magnetic force on the wire must balance the weight of the wire: \[ \text{Magnetic force} = \text{Weight}. \] The magnetic force on a current-carrying wire is \(F_B = I \,\ell\,B\). The weight of the wire is \(mg\). Hence: \[ I \,\ell\,B = mg. \]
3. Substitute the Values:
   \[ I \times (1.0\,\text{m}) \times (0.2\,\text{T}) = 0.1\,\text{kg} \times 10\,\text{m/s}^2. \] Which simplifies to: \[ 0.2\,I = 1.0. \] Thus: \[ I = \frac{1.0}{0.2} = 5\,\text{A}. \]
4. Final Answer: The current needed is \(5\,\text{A}\), matching choice (3).

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