Tag: magnetic field on the axis of a toroid

Questions Related to magnetic field on the axis of a toroid

A beam of protons with a velocity $4 \times 10^5 ms^{-1}$ enters a uniform magnetic field of 0.3 T at an angle of $60^o$ to the magnetic field. Find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of the rotation). Mass of the proton $= 1.67 \times 10^{-27} kg$

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

  1. 2.3 cm

  2. 5.35 cm

  3. 4.35 cm

  4. 6.35 cm

Show answer
Correct Option: C
Explanation:

When a charged particle is projected at an angle $\theta$ to a magnetic field, the component of velocity parallel to the field is $v\cos \theta$ while perpendicular to the field is $v\sin \theta$, so the particle will move in a circle of radius
$r = \dfrac{m(v\sin \theta)}{qB}=\dfrac{( 1.67 \times 10^{-27}) \times \left( 4 \times 10^{5} \times \sin 60^0 \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{ \left( 1.67 \times 10^{-27} \right) \times \left( 4 \times 10^{5} \times \dfrac{\sqrt{3}}{2} \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{2 \times 10^{-2}}{\sqrt{3}}$
Time period: $T = \dfrac{2\pi r}{vsin\theta}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \sin 60^0}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \dfrac{\sqrt{3}}{2}}=\dfrac{2\pi}{3} \times 10^{-7}$
Pitch: $P = v \cos \theta T= (4\times 10^5) \times \cos 60^0 \times \dfrac{2\pi}{3} \times 10^{-7}=\dfrac{4\pi}{3}\times 10^{-2}= 4.35 \times 10^{-2}: m$

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

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

  1. 1000

  2. 3000

  3. 5000

  4. 10000

Show answer
Correct Option: C
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$

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?

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

  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. $1.2 \times 10^{4} kg-m^2$

Show answer
Correct Option: D
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}$

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

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