Question

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

Answer 1

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

           F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction       f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

 B_w = B sinθ

 B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

       B = 0.1 T