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MCQEasyJEE 2023Force on Current-Carrying Conductor

JEE Physics 2023 Question with Solution

A straight wire AB of mass 40 g40 \, g and length 50 cm50 \, cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T0.40 \, T. The magnitude of the current required in the wire to remove the tension in the supporting leads is _____ A. (Take g=10 m/s2g = 10 \, m/s^2)

  • A

    1.5 A1.5 \, A

  • B

    2 A2 \, A

  • C

    3 A3 \, A

  • D

    4 A4 \, A

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: mass of the wire m=40 g=0.040 kgm = 40 \, g = 0.040 \, kg, length L=50 cm=0.5 mL = 50 \, cm = 0.5 \, m, magnetic field B=0.40 TB = 0.40 \, T, and g=10 m/s2g = 10 \, m/s^2.

Find: the current II required so that the tension in the supporting leads becomes zero.

For equilibrium, the magnetic force on the wire must balance its weight.

Fmagnetic=FgravityF_{\text{magnetic}} = F_{\text{gravity}}

The magnetic force is

Fmagnetic=ILBF_{\text{magnetic}} = I L B

The gravitational force is

Fgravity=mgF_{\text{gravity}} = mg

Equating the two forces,

ILB=mgI L B = mg

So,

I=mgLBI = \frac{mg}{LB}

Substitute the values,

I=0.040⋅100.5⋅0.40I = \frac{0.040 \cdot 10}{0.5 \cdot 0.40}I=0.400.2=2 AI = \frac{0.40}{0.2} = 2 \, \text{A}

Therefore, the required current is 2 A2 \, \text{A}. The correct option is B.

Force Balance Interpretation

Given: the wire is suspended by flexible leads in a uniform magnetic field.

Find: the current at which the leads have no tension.

If the supporting leads have zero tension, they are no longer supporting the weight of the wire. Hence the upward magnetic force alone must equal the downward weight of the wire.

Thus use

ILB=mgI L B = mg

Now convert all quantities into SI units before substitution:

  • m=40 g=0.040 kgm = 40 \, g = 0.040 \, kg
  • L=50 cm=0.5 mL = 50 \, cm = 0.5 \, m
  • B=0.40 TB = 0.40 \, T

Then,

I=0.040×100.5×0.40I = \frac{0.040 \times 10}{0.5 \times 0.40}I=0.400.20=2I = \frac{0.40}{0.20} = 2

Therefore, the current needed is 2 A2 \, \text{A}.

Common mistakes

  • Using mass as 40 kg40 \, kg instead of converting 40 g40 \, g to 0.040 kg0.040 \, kg is incorrect because SI units are required in the formula. Always convert grams to kilograms before substituting.

  • Using the wire length as 50 m50 \, m or 50 cm50 \, cm directly in the equation is incorrect because LL must be in metres. Use 50 cm=0.5 m50 \, cm = 0.5 \, m.

  • Forgetting that zero tension means the magnetic force fully balances the weight leads to an incomplete force balance. Set ILB=mgILB = mg, not magnetic force plus tension equals weight.

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