Tag: magnetic field due to a straight current carrying conductor

Questions Related to magnetic field due to a straight current carrying conductor

A particle of charge per unit mass $\alpha$  is released from origin with a velocity $\mathop v\limits^ \to   = {v _0}\mathop i\limits^ \wedge  $ in a uniform magnetic field $\mathop B\limits^ \to   =  - {B _0}\mathop k\limits^ \wedge  $ . If the particle passes through (0,y,0) then y is
equal to

  1. $ - \frac{{2{v _0}}}{{{B _0}\alpha }}\,\;$

  2. $\frac{{{v _0}}}{{{B _0}\alpha }}\;$

  3. $\;\frac{{2{v _0}}}{{{B _0}\alpha }}$

  4. $ - \frac{{{v _0}}}{{{B _0}\alpha }}$


Correct Option: C

A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it at a small distance and another point Q lies to the west at the same distance the magnetic field at P is :

  1. greater than at Q

  2. same as at Q

  3. less than at Q

  4. greater or less than at Q depending upon the strength of current


Correct Option: B
Explanation:

As per Biot Savart Law, Magnetic field at a point is inversely proportional to square of distance from the current carrying conductor. Therefore magnitude of magnetic field is same at both points P and Q, irrespective of their position from the conductor.

If a current carrying wire carries 10A current then the magnetic field is X. Now the current in the wire increases to 100A, them magnetic field in the wire becomes

  1. >X

  2. <X

  3. =X

  4. all


Correct Option: A
Explanation:

The Magnitude of magnetic field produced by a straight current carrying wire at a given point is
A) directly proportional to the current passing in the wire and
B) inversely proportional to the distance of that point from the wire.

The magnetic field produced by a current-carrying wire at a given point depends on

  1. the current passing through it.

  2. the voltage across it

  3. the power through it

  4. all


Correct Option: A
Explanation:

The magnetic field produced by a current-carrying wire at a given point depends directly on the current passing through it. Therefore, if there is a circular coil having n turns, the field produced is n times as large as that produced by a single turn. This is because the current in each circular turn has the same direction, and the field due to each turn then just adds up.

If magnetic field produced by a straight current carrying wire at a distance 10cm from it is X. Then the magnetic field produced at a distance 29cm will be

  1. >X

  2. <X

  3. =X

  4. all


Correct Option: B
Explanation:

The Magnitude of magnetic field produced by a straight current carrying wire at a given point is 
A) directly proportional to the current passing in the wire and 
B) inversely proportional to the distance of that point from the wire.

Which of the following relation represents Biot-Savart's law?

  1. $\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ r } $

  2. $\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \hat { r } }{ { r }^{ 3 } } $

  3. $\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 3 } } $

  4. $\vec { dB } =\dfrac { { \mu } _{ 0 } }{ 4\pi } \dfrac { \vec { dl } \times \vec { r } }{ { r }^{ 4 } } $


Correct Option: C
Explanation:

If unit current is flowing through the conductor, then Biot-Savart's law is represented as
    $\vec { dB } =\dfrac { { \mu  } _{ 0 } }{ 4\pi  } \dfrac { \vec { dl } \times \vec { r }  }{ { r }^{ 3 } } $

A long cylindrical wire of radius R carries a current $i$ distributed uniformly over its cross section.Find the maximum magnetic field produced by this wire.

  1. $\dfrac{\mu _0 il}{2\pi R}$

  2. $\dfrac{8\mu _0l}{\pi R}$

  3. $\dfrac{\mu _0l}{4\pi R}$

  4. $\dfrac{2\mu _0l}{\pi R}$


Correct Option: A
Explanation:

Let the current $i$ is uniformly distributed throughout the wire.

$R$ is the radius of the cylindrical wire.
From Ampere's circuital law, $\phi B\times dl=B\times 2\pi l$ (where $l$ is a small part of the ring)
For maximum magnetic field, $B={ \mu  } _{ 0 }\dfrac { i{ l }^{ 2 } }{ R } \times \dfrac { 1 }{ 2\pi l } \ \quad \quad =\dfrac { { \mu  } _{ o }il }{ 2\pi R } $
 

Motion of charges is noting but :

  1. Electric current

  2. magnetic effect

  3. heating effect

  4. all of the above


Correct Option: D
Explanation:
Due to motion of charges through a conductor, the current of charges passing through its cross-section changes. It causes electric current $\left( ie. \dfrac{da}{dt}\right)$. Because of this current, heating effect and magnetic effect are produced in accordance with Joules and Bio Savaert's law resp.
Option $-D$ is correct.

A condenser is charged using a constant current. The ratio of the magnetic field at a distance of R/2 and R from the axis is (R the radius of plate)

  1. 1:1

  2. 2:1

  3. 1:2

  4. 1:4


Correct Option: C

The magnetic field at the origin due to a current element $i.\vec {dl}$ placed at a position $\vec r$ is

  1. $\dfrac {\mu _0i}{4\pi} \dfrac {\vec {dl}\times \vec r}{r^3}$

  2. $\dfrac {\mu _0i}{4\pi} \dfrac {\vec r\times \vec {dl}}{r^3}$

  3. $-\dfrac {\mu _0i}{4\pi} \dfrac {\vec r\times \vec {dl}}{r^3}$

  4. $-\dfrac {\mu _0i}{4\pi} \dfrac {\vec {dl}\times \vec r}{r^3}$


Correct Option: A,C
Explanation:

$B= \dfrac{\mu _0}{4\pi}i\dfrac{\vec{dl}\times  \vec{r}}{r^3}=-\dfrac{\mu _0}{4\pi}i\dfrac{\vec{r}\times  \vec{dl}}{r^3}$
since $\vec{dl}\times  \vec{r} = -(\vec{r}\times  \vec{dl})$
Options (A) and (C) are correct