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MCQEasyJEE 2024Force on Current-Carrying Conductor

JEE Physics 2024 Question with Solution

The horizontal component of Earth’s magnetic field at a place is 3.5×10−5 T3.5 \times 10^{-5} \, \text{T}. A very long straight conductor carrying a current of 2 A\sqrt{2} \, \text{A} is placed from South East to North West. The force per unit length experienced by the conductor is:

  • A

    35×10−6 N/m35 \times 10^{-6} \, \text{N/m}

  • B

    50×10−6 N/m50 \times 10^{-6} \, \text{N/m}

  • C

    25×10−6 N/m25 \times 10^{-6} \, \text{N/m}

  • D

    10×10−6 N/m10 \times 10^{-6} \, \text{N/m}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Horizontal component of Earth’s magnetic field is BH=3.5×10−5 TB_H = 3.5 \times 10^{-5} \, \text{T} and current is I=2 AI = \sqrt{2} \, \text{A}.

Find: Force per unit length on the conductor.

The magnetic force on a straight conductor is given by

F⃗=I(L⃗×B⃗)\vec{F} = I(\vec{L} \times \vec{B})

So, the magnitude of force per unit length is

FL=IBsin⁡θ\frac{F}{L} = I B \sin \theta

The horizontal component of Earth’s magnetic field is directed from South to North. The conductor is placed from South East to North West, so the angle between current direction and magnetic field is 45∘45^\circ.

Substituting the values,

FL=(2)(3.5×10−5)sin⁡45∘\frac{F}{L} = (\sqrt{2})(3.5 \times 10^{-5})\sin 45^\circ

Using

sin⁡45∘=12\sin 45^\circ = \frac{1}{\sqrt{2}}

we get

FL=(2)(3.5×10−5)(12)\frac{F}{L} = (\sqrt{2})(3.5 \times 10^{-5})\left(\frac{1}{\sqrt{2}}\right) FL=3.5×10−5 N/m\frac{F}{L} = 3.5 \times 10^{-5} \, \text{N/m}

Now,

3.5×10−5=35×10−63.5 \times 10^{-5} = 35 \times 10^{-6}

Therefore, the force per unit length experienced by the conductor is 35×10−6 N/m35 \times 10^{-6} \, \text{N/m}. The correct option is A.

Direct Cancellation Trick

Given: I=2 AI = \sqrt{2} \, \text{A}, BH=3.5×10−5 TB_H = 3.5 \times 10^{-5} \, \text{T}, and angle θ=45∘\theta = 45^\circ.

Find: FL\frac{F}{L}.

Use

FL=IBsin⁡θ\frac{F}{L} = I B \sin \theta

Since

I=2,sin⁡45∘=12I = \sqrt{2}, \qquad \sin 45^\circ = \frac{1}{\sqrt{2}}

the two factors cancel immediately.

So,

FL=3.5×10−5 N/m=35×10−6 N/m\frac{F}{L} = 3.5 \times 10^{-5} \, \text{N/m} = 35 \times 10^{-6} \, \text{N/m}

Therefore, the correct option is A.

Common mistakes

  • Using the full Earth’s magnetic field instead of the horizontal component is incorrect because the question explicitly gives BHB_H. The force here must be calculated with BH=3.5×10−5 TB_H = 3.5 \times 10^{-5} \, \text{T}.

  • Taking the angle as 135∘135^\circ or 90∘90^\circ is incorrect. The conductor is along South East to North West, while B⃗H\vec{B}_H is along South to North, so the acute angle between them is 45∘45^\circ.

  • Forgetting that sin⁡45∘=12\sin 45^\circ = \frac{1}{\sqrt{2}} leads to a wrong numerical result. Substitute the trigonometric value before simplifying.

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