There are 50 turns per cm length in a very lo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnetic field $B$ at the center of a long solenoid is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length a

Vedclass · Accepted Answer

$5 \pi 	imes 10^{-3}$

Vedclass · Answer

(B) The magnetic field $B$ at the center of a long solenoid is given by the formula $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.Given:Number of turns per unit length $n = 50 	ext{ turns/cm} = 50 	imes 10^2 	ext{ turns/m} = 5000 	ext{ turns/m}$.Current $I = 2.5 \ A$.Permeability of free space $\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$.Substituting the values:$B = (4 \pi 	imes 10^{-7}) 	imes (5000) 	imes (2.5)$$B = 4 \pi 	imes 10^{-7} 	imes 12500$$B = 4 \pi 	imes 1.25 	imes 10^{-3}$$B = 5 \pi 	imes 10^{-3} \ T$.

There are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.

Similar Questions

State Ampere's circuital law and explain its mathematical expression.

$A$ thin hollow copper pipe carries a direct current. Which of the following statements is incorrect?

The magnetic energy stored in a long solenoid of area of cross-section $A$ in a small region of length $L$ is

$A$ current-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity,then the acceleration $(a)$ of the charged particle will satisfy:

Vedclass Test Series

Exam Paper Generator

Online Exam Module