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JEE Mains Previous Paper 1 (Held On: 10 Jan 2019 Shift 2)

Option 1 : 6.5 × 10^{-5} N

**Concept:**

Without applied forces, (in equilibrium position) the needle will stay in the resultant magnetic field of earth. Hence, the dip ‘θ’ at this place is 45° (given).

We know that, horizontal and vertical components of earth’s magnetic field (B_{H} and B_{v}) are related as

\(\frac{{{B_v}}}{{{B_H}}} = tan\;\theta\)

**Calculation:**

Given,

The horizontal component of earth's magnetic field B_{H} is 18 × 10^{-6} T

Length of the magnetic needle = 0.12 m

Strength of the pole, m = 1.8 A-m

Here, θ = 45° and B_{H} = 18 × 10^{-6 }T

B_{v} = B_{H} tan 45°

B_{v} = B_{H}

B_{v} = B_{H} = 18 × 10^{-6 }T (∵ tan 45° = 1)

Now, when the external force F is applied, so as to keep the needle stays in horizontal position is shown below,

Taking torque at point P, we get

mB_{v} × 2l = Fl

∴ F = 2 × mB_{v}

Substituting the given values, we get,

F = 2 × 1.8 × 18 × 10^{-6}

F = 64.8 × 10^{-6}

F = 6.48 × 10^{-5} = 6.5 × 10^{-5 }N

Therefore, the vertical force that should be applied at one of its ends to keep the needle horizontal is 6.5 × 10^{-5 }N.

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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