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JEE Main 202325 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2023 · 25 January · Morning Shift · Physics · Magnetic Effects of Current

A solenoid of 1200 turns is wound uniformly in a single layer on a...

A solenoid of 12001200 turns is wound uniformly in a single layer on a glass tube 2m2m long and 0.2m0 . 2m in diameter. The magnetic intensity at the center of the solenoid when a current of 2A2A flows through it is:

Options

  1. A2.4×103Am−12 . 4 \times {10}^{3}A{m}^{- 1}
  2. B1.2×103Am−11 . 2 \times {10}^{3}A{m}^{- 1}
  3. C1Am−11A{m}^{- 1}
  4. D2.4×10−3Am−12 . 4 \times {10}^{- 3}A{m}^{- 1}

Correct answer

B. 1.2×103Am−11 . 2 \times {10}^{3}A{m}^{- 1}

Step-by-step solution

Given,

Number of turns per metre n=12002n = \dfrac{1200}{2}

l=2m,d=0.2mandI=2Al = 2m, d = 0 . 2m\text{and}I = 2A

Magnetic field at centre inside the solenoid is given by B=μ0nIB = {\mu}_{0}nI, where, μ0{\mu}_{0} is the permeability constant.

Magnetic intensity is related to magnetic field by

H=Bμ0=μ0nIμ0=nI=(12002)(2)H = \dfrac{B}{{\mu}_{0}} = \dfrac{{\mu}_{0}nI}{{\mu}_{0}} = nI = \left(\dfrac{1200}{2}\right) ( 2 )

⇒H=1.2×103Am−1⇒ H = 1 . 2 \times {10}^{3}A{m}^{- 1}

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