A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:

Correct answer: B. 1.2 × 10^3 Am^ -1 Given that number of turns, n=1200 Current, I=2 A Magnetic field at centre inside the solenoid is given by, B=μ_0 nI So magnetic intensity at centre of the solenoid, aligned & H= B μ_0 =nI=((1200)/(2))(2) ( B=μ_0 H) & H=1.2 × 10^3 Am^ -1 aligned

BITSAT 2023Paper 2

Exam: BITSAT
Year: 2023
Paper: Paper 2
Subject: Physics
Chapter: Magnetic Effects of Current

BITSAT 2023 · Paper 2 · Physics · Magnetic Effects of Current

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

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

2 \mathrm{ m}

long and

0.2 \mathrm{ m}

in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:

Options

A $2.4 \times 10^3 \mathrm{Am}^{-1}$
B $1.2 \times 10^3 \mathrm{Am}^{-1}$
C $1 \mathrm{Am}^{-1}$
D $2.4 \times 10^{-3} \mathrm{ A} \mathrm{ m}^{-1}$

Correct answer

B. $1.2 \times 10^3 \mathrm{Am}^{-1}$

Step-by-step solution

Given that number of turns,

n=1200

Current,

\mathrm{I}=2 \mathrm{ A}

Magnetic field at centre inside the solenoid is given by,

\mathrm{B}=\mu_0 \mathrm{nI}

So magnetic intensity at centre of the solenoid,

\begin{aligned} \\ & \mathrm{H}=\frac{\mathrm{B}}{\mu_0}=\mathrm{nI}=\left(\frac{1200}{2}\right)(2) \quad\left(\because \mathrm{B}=\mu_0 \mathrm{H}\right) \\ \\ & \mathrm{H}=1.2 \times 10^3 \mathrm{Am}^{-1} \\ \end{aligned}

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