In gravity free space, a bead of charge 1 μ C and mass 3 mg is threaded on a rough rod of friction coefficient μ =0.3. A magnetic field of magnitude 0.2 T exists perpendicular to the rod. The bead is projected along the rod with a speed of 4 m/s . How much distance (in m ) will the bead cover before coming to rest?

Question

In gravity free space, a bead of charge 1 μ C and mass 3 mg is threaded on a rough rod of friction coefficient μ =0.3. A magnetic field of magnitude 0.2 T exists perpendicular to the rod. The bead is projected along the rod with a speed of 4 m/s . How much distance (in m ) will the bead cover before coming to rest? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-lmudavmj5d9dsd7f.jpg" /> N=qvB -μ qvB=(m d v/d t)=mv(d v/d S) μ qBS=mv S=(3 × 10- 6 × 4/0 . 3 × 10- 6 × 0 . 2)=200 m

N=qvB −μqvB=dtmdv​=mvdSdv​ μqBS=mv S=0.3×10−6×0.23×10−6×4​=200m

$N = q v B$
$- μ q v B = \frac{m d v}{d t} = m v \frac{d v}{d S}$
$μ qBS = m v$
$S = \frac{3 \times 1 0 ^{- 6} \times 4}{0.3 \times 1 0 ^{- 6} \times 0.2} = 200 m$