Tag: magnetic field lines due to current

Questions Related to magnetic field lines due to current

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

A proton is moving with velocity ${10}^{4}m/s$ parallel to the magentic field of intensity 5 tesla.The force on the proton is

  1. $8\times {10}^{-15}N$
  2. ${10}^{4}N$
  3. $1.6\times {10}^{-19}N$
  4. Zero

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

Answer is D.

The magnitude and direction of F depend on the velocity of the particle and on the magnitude and direction of the magnetic field B.
When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero.
When the particles velocity vector makes any angle with the magneticfield, the magnetic force acts in a direction perpendicular to both v and B; that is, F is perpendicular to the plane formed by v and B.
The magnitude of the magnetic force is
$F=qvBsin\theta $
where $\theta $ is the smaller angle between v and B. From this expression, we see that F is zero when v is parallel or antiparallel to B or 180) and maximum when v is perpendicular to B.
In this case, as the proton moves parallel to the magnetic field, the force is zero.

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

The pattern of the magnetic field around a conductor due to an electric current flowing through it depends on

  1. amount of current flowing through the conductor

  2. amount of voltage supplied to the conductor

  3. size of conductor

  4. shape of the conductor

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know that;
$\vec B=\dfrac{\mu _0}{4\pi}\int { \dfrac { Idl\times {\vec  r  } }{ \left| { r }^{ 1 } \right| ^{ 3 } }  } $
Where, $\vec B$ is magnetic field, $\vec I$ is current.
Thus magnitude of $\vec B$ depend on I, but pattern of magnetic field depend on shape of conductor as the direction of magnetic field is obtained used Cut finger rule i.e, pointing right hand thumb in current direction, Curl fingers describes direction thus pattern.
Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

The value of $\mu$ is $4 \pi \times {10}^{-7} H {m}^{-1}$.

  1. True

  2. False

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

The physical constant $μ _0$ commonly called the vacuum permeability, permeability of free space, ... In the reference medium of classical vacuum, μ0 has an exact defined value: .... The value of $μ _0$ was chosen such that the rmks unit of current is equal in size to the ampere in the emu system: μ0 is defined to be $4π × 10^{−7} H/m.$

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

The value of magnetic field due to a small element of current carrying conductor at a distance r and lying on the plane perpendicular to the element of conductor is

  1. Zero

  2. Maximum

  3. Inversely proportional to the current

  4. None of the above

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

The Biot Savart law is used for computing the resultant magnetic field B at position r generated by a steady current I.  According to it $B=\frac {\mu _0 I } {4\pi} \int \frac {I\vec{dl} \times \vec{r}} { r^3}$.
So magnetic field at a distance r and lying on the plane perpendicular ($\pi /2$)to the element of conductor is maximum .

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

The magnetic field due to current flowing in a ling straight conductor is directly proportional to the current and inversely proportional to the distance of the point of observation from the conductor. What is this law known as?

  1. Blonde-Rey law

  2. Biot-Savart's law

  3. Beer-Lambert law

  4. Ampere's law

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

The magnetic field due to a long straight conductor is given by the Biot-Savart law, which leads to the formula B = mu0*i / (2*pi*r).

Multiple choice physics moving charges and magnetism field due to a current carrying conductor magnetic field due to a straight current carrying conductor magnetic field lines due to current

A current of i ampere is flowing in an equilateral triangle of side a. The magnetic induction at the centroid will be?

  1. $\dfrac{\mu _i}{3\sqrt{3}\pi a}$
  2. $\dfrac{3\mu _i}{2\pi a}$
  3. $\dfrac{5\sqrt{2}\mu _i}{3\pi a}$
  4. $\dfrac{9\mu _i}{2\pi a}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
a = length of side of equilateral triangle 

r = perpendicular distance of each side from centroid $=\frac { \sqrt { 3a }  }{ 6 } $

θ = angle by each end of each side at centroid = 60

Using Biot-savart's law , magnetic field at the centroid by each side is given as 

$B=\frac { μ }{ 4π } \ast \frac { i }{ r } \ast (Sinθ+Sinθ)$

$B=\frac { μ }{ 4π } \left( \frac { i }{ \sqrt { 3 }  } \frac { a }{ 6 }  \right) (Sin60+Sin60)$

$B=\frac { μ }{ 4π } \left( \frac { 6i }{ \sqrt { 3a }  } \frac { \sqrt { 3 }  }{ 2 }  \right) $

$B=\frac { μ }{ 4π } \frac { 6i }{ a } $

total magnetic field by all three sides is given as 

$B''=3B=3\frac { μ }{ 4π } \frac { 6i }{ a } $

$B''=\frac { μ }{ 2π } \frac { 9i }{ a } $




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 carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of the magnetic field is

  1. 4 B

  2. B/2

  3. B

  4. 2 B

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

$\displaystyle B = \mu _0 N _0  i;   $

$B _1 = (\mu _0) \left ( \dfrac{N _0}{2} \right ) (2 i) $

$= \mu _0 N _0 i = B$


$\Rightarrow B _1 = B$

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 wire 28 m long is bent into N turns of circular coil of diameter 14 cm forming a solenoid of length 60 cm. Calculate the magnetic field inside it when a current of 5 amp passed through it.  $(\mu _0 = 12.57 \times 10^{-7} m^{-1})$

  1. $6.67 \times10^{-1} T$
  2. $6.67 \times10^{-4} T$
  3. $6.67 \times10^{4} T$
  4. $2.67 \times10^{-4} T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $d = 14 cm=0.14 m$    

           $l = 60cm = 0.6 m$

By the question, $N \times \pi d = 28 m.$

$N \times \pi \times 0.14 = 28$

$\displaystyle \therefore N = \dfrac{28}{0.14 \times \pi} = 63.66 turns$

$\displaystyle B = \mu _o   nI  =\mu _o  \dfrac{N}{l} I = 12.57 \times 10^{-7} \times \dfrac{63.66}{0.6} \times 5$

$=6.67 \times10^{-4} T$

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

The electric current in a circular coil of two turns produced a magnetic induction of $0.2 T$ at its centre. The coil is unwound and is rewound into a circular coil of four turns. The magnetic induction at the centre of the coil now is, in $T$ :
(if same current flows in the coil)

  1. $0.2$
  2. $0.4$
  3. $0.6$
  4. $0.8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Coil is unwound and is rewound into a circular coil of 4 turns,
$\therefore $ $2\pi R=4\times 2\pi r$ 
$r=\dfrac{R}{4}$
$\left | B \right | =\dfrac{\mu _{0}I}{\dfrac{2R}{4}}$ $=0.2\times$ $4=0.8T$