A thin conducting rod of rectangular cross section and of length 'ℓ'=25 meter and mass ‘m’ = 4 kg lies on the horizontal table. It can translate along x-axis. Coefficient of friction between the rod and the table is 'μ'=12. If the current in the conductor is 2 ampere, then the minimum magnitude of uniform constant magnetic field strength (in Tesla) that should exist in space such that conducting rod just starts to translate along x-axis is P .Value of P= ------- (take g = 10 m/sec2)

Let say magnetic field acts such that magnetic force F makes an angle θ with horizontal.Vertical : N=mg-FsinθHorizontal : fL=Fcosθ⇒μN=Fcosθ⇒μ(mg-Fsinθ)=Fcosθ⇒F=μmgμsinθ+cosθFor F to be minimum, (μsinθ+cosθ) should be maximum∴ddθ(μsinθ+cosθ)=0⇒μcosθ−sinθ=0⇒μ=sinθcosθ⇒sinθ=μ1+μ2 and cosθ=11+μ2 ∴Fmin=μmg1+μ2⇒iℓBmin=μmg1+μ2⇒Bmin=μmgiℓ1+μ2=(0.5)(40)(45)(52)=2T

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A thin conducting rod of rectangular cross section and of length $' ℓ' = 2 \sqrt{5}$ meter and mass ‘m’ = 4 kg lies on the horizontal table. It can translate along x-axis. Coefficient of friction between the rod and the table is $' μ' = (\frac{1}{2})$ . If the current in the conductor is 2 ampere, then the minimum magnitude of uniform constant magnetic field strength (in Tesla) that should exist in space such that conducting rod just starts to translate along x-axis is P .Value of P= ------- (take
g = 10 $m / \sec^{2}$ )

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Detailed Solution

$\begin{array}{l} Let say magnetic field acts such that magnetic force F makes an angle θ with horizontal. \\ Vertical : N=mg-Fsin θ \\ {Horizontal : f}_{L} = F \cos θ \\ \Rightarrow μ N = F \cos θ \\ \Rightarrow μ (mg-Fsin θ) = F \cos θ \\ \Rightarrow F = \frac{μ m g}{μ \sin θ + \cos θ} \\ For F to be minimum, (μ \sin θ + \cos θ) should be maximum \\ ∴ \frac{d}{d θ} (μ \sin θ + \cos θ) = 0 \\ \Rightarrow μ \cos θ - \sin θ = 0 \\ \Rightarrow μ = \frac{\sin θ}{\cos θ} \\ \Rightarrow \sin θ = \frac{μ}{\sqrt{1 + μ^{2}}} a n d \cos θ = \frac{1}{\sqrt{1 + μ^{2}}} \\ ∴ F_{\min} = \frac{μ m g}{\sqrt{1 + μ^{2}}} \\ \Rightarrow i ℓ B_{\min} = \frac{μ m g}{\sqrt{1 + μ^{2}}} \\ \Rightarrow B_{\min} = \frac{μ m g}{i ℓ \sqrt{1 + μ^{2}}} = \frac{(0.5) (40)}{(4 \sqrt{5}) (\frac{\sqrt{5}}{2})} = 2 T \end{array}$