An ideal solenoid having 5000 turns / m has an aluminium core and carries a current of 5 A . If _ Al =2.3 × 10^ -5 , then the magnetic field developed at centre will be

Correct answer: A. 0.031 T Given, n=5000 turns / m, i=5 A, _ Al =2.3 × 10^ -5 , B= ? As, B=μ_0(H+I) where, aligned H & =(B_0)/(μ_0)=(μ_0 n i)/(μ_0)=n i & =5000 × 5=2.5 × 10^4 A / m aligned aligned & and I= H=2.3 × 10^ -5 × 2.5 × 10^4 & =0.575 A / m & B=μ_0(H+l) & =4 π × 10^ -7 (2.5 × 10^4+0.575) T & =0.031 T & aligned

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Magnetic Effects of Current

JIPMER 2016 · Physics · Magnetic Effects of Current

An ideal solenoid having 5000 turns / m has an aluminium core and...

An ideal solenoid having 5000 turns

/ \mathrm{m}

has an aluminium core and carries a current of 5 A . If

\chi_{\mathrm{Al}}=2.3 \times 10^{-5}

, then the magnetic field developed at centre will be

Options

A0.031 T
B0.048 T
C0.027 T
D0.050 T

Correct answer

A. 0.031 T

Step-by-step solution

Given,

n=5000

turns

/ \mathrm{m}, \quad i=5 \mathrm{ A}, \quad \chi_{\mathrm{Al}}=2.3 \times 10^{-5}

B=

? As,

\quad B=\mu_0(H+I)

where,

\quad

\begin{aligned} H & =\frac{B_0}{\mu_0}=\frac{\mu_0 n i}{\mu_0}=n i \\ & =5000 \times 5=2.5 \times 10^4 \mathrm{ A} / \mathrm{m}\end{aligned}

\begin{aligned} & \text{\text{ and } } \quad I=\chi H=2.3 \times 10^{-5} \times 2.5 \times 10^4 \\ & =0.575 \mathrm{ A} / \mathrm{m} \\ & \therefore \quad B=\mu_0(H+l) \\ & =4 \pi \times 10^{-7}\left(2.5 \times 10^4+0.575\right) \mathrm{T} \\ & =0.031 \mathrm{ T} \\ & \end{aligned}

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