Tag: magnetic field lines due to current

Questions Related to magnetic field lines due to current

Multiple choice physics magnetic effect of electric current magnetic field lines due to current magnetic field due to current carrying conductor magnetic field on the axis of a toroid

A long solenoid has magnetic field strength of $3.14\times 10^{-2}\ T$ inside it when a current of $5\ A$ passes through it. The number of turns in $1\ m$ of the solenoid is

  1. 1000

  2. 3000

  3. 5000

  4. 10000

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

No of turns per unit length for an infinite  solenoid: $n =

\dfrac{B}{\mu _0 i}=\dfrac{3.14 \times 10^{-2}}{4\pi \times 10^{-7} \times 5}=5000$

Multiple choice physics magnetic effect of electric current magnetic field lines due to current magnetic field due to current carrying conductor magnetic field on the axis of a toroid

A circular coil of $16$ turns and radius $10$ cm carrying a current of $0.75 A$ rests with its plane normal to an external field of magnitude $5.0 \times 10^{-2}T$. The coil is free to turn about an axis in its plane perpendicular to the filed direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0/s$. What is the moment of inertia of the coil about its axis of rotation?

  1. $1.2 \times 10^{4} g-cm^2$
  2. $3\times 10^{4} kg-m^2$
  3. $0.3 \times 10^{4} kg-m^2$
  4. $1.2 \times 10^{4} kg-m^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,
Number of turns in the circular coil, $N = 16$
Radius of the coil, $r = 10$ cm $= 0.1m$ 
Current in the coil, $I = 0.75A$ 
Magnetic field strength, $B = 5.0 \times 10^{2}T$ 
Frequency of oscillations of the coil, $v = 2.0s^{-1}$
Magnetic moment, $M = NIA$
$M = NI\pi r^2 = (16)(0.75)\pi(0.1)^2 = 0.3768 JT^{-1}$
The frequency is given by,
$ \nu = \dfrac{1}{2\pi}\sqrt {\dfrac{MB}{I}}$

$ I = \dfrac{MB}{4 \pi^2 \nu^2}$

$I = \dfrac{(0.3768)(5 \times 10^{2})}{4 \pi^2 2^2}$

$I = 1.19 \times 10^{4} = 1.2 \times 10^{4} Kg-m^{2}$