Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Magnetic Properties of Matter

NTA Abhyas 2020 · Physics · Magnetic Properties of Matter

A wire of length L = 20cm ~is bent into a semi-circular arc and the...

A wire of length L=20cmL = 20cm~is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges +Q+ Q~and& −Q- Q~as shown in the figure. The magnitude of the charge on each half is& \left|Q\right| = {10}^{3}{ε}_{0},where, where{ε}_{0}~is the permittivity of free the space. The net electric field at the centre&O$& is

A wire of length L = 20cm ~is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged

Options

  1. A\left(25 \times {10}^{3}\right)\overset{^}{i}{NC}^{- 1}
  2. B\left(50 \times {10}^{3}\right)\overset{^}{i}{NC}^{- 1}
  3. C\left(25 \times {10}^{3}\right)\overset{^}{j}{NC}^{- 1}
  4. D\left(50 \times {10}^{3}\right)\overset{^}{j}{NC}^{- 1}

Correct answer

A. \left(25 \times {10}^{3}\right)\overset{^}{i}{NC}^{- 1}

Step-by-step solution

A wire of length L = 20cm ~is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged

L=\piRL = \piR

⇒R=Lπ=20100πm=15πm⇒ R = \dfrac{L}{π} = \dfrac{20}{100\pi} \text{m} = \dfrac{1}{5\pi} \text{m}

⇒⇒ due to a charge arc, electric field at centre is given by

E=2KλRsinθ2E = \dfrac{2K\lambda}{R} sin\dfrac{θ}{2}

A wire of length L = 20cm ~is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged

E1=E2=2k⋋Rsin902⋋=Q\piR/2{E}_{1} = {E}_{2} = \dfrac{2k ⋋}{R}sin \dfrac{90}{2} { ⋋ = \dfrac{Q}{\piR / 2} }

E1=E2=22KQ\piR2{E}_{1} = E{}_{2} = \dfrac{2\sqrt{2} KQ}{\piR^{2}}

Component along \overset{^}{j} gets cancelled and

Enet=2E1{E}_{net} = \sqrt{2} {E}_{1}

=4KQ\piR2= \dfrac{4KQ}{\piR^{2}}

=4×1(103ϵ0)4πϵ0\piR2=4×103ϵ04π2ϵ0R2=102R2=100(15π)2=25×103= \dfrac{4 \times 1\left({10}^{3}{ϵ}_{0}\right)}{4\pi{ϵ}_{0}\piR^{2}} = \dfrac{4 \times {10}^{3}{ϵ}_{0}}{4{\pi}^{2}{ϵ}_{0}R^{2}} = \dfrac{{10}^{2}}{R^{2}} = \dfrac{100}{{\left(\dfrac{1}{5\pi}\right)}^{2}} = 25 \times {10}^{3}

{\overset{→}{E}}_{net} = 25 \times {10}^{3}\dfrac{N}{C}\overset{^}{i}
Practice Magnetic Properties of Matter on Quantrex Academy →

More from Magnetic Properties of Matter

A solenoid has a core made of material with relative permeability 400 . The magnetic… JEE Main 2026Two identical small bar magnets each of dipole moment 3 5 J/T are placed at a center to… JEE Main 2026Given below are two statements: one is labelled as Assertion (A) and the other is… JEE Main 2026Given below are two statements : one is labelled as Assertion (A) and the other is… JEE Main 2025The percentage increase in magnetic field (B) when space within a current carrying… JEE Main 2025A solenoid having area A and length ' l ' is filled with a material having relative… JEE Main 2025The relationship between the magnetic susceptibility (χ) and the magnetic permeability… JEE Main 2025A bar magnet has total length 2 l=20 units and the field point P is at a distance d=10… JEE Main 2025 Full Magnetic Properties of Matter list All NTA Abhyas JEE Main PYQs