Tag: magnetic effect of electric current

Questions Related to magnetic effect of electric current

A small coil of N turns has an area A and a current i flows through it. The magnetic dipole moment of the coil will be

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. i NA

  2. $i^{2}$ NA

  3. i $N^{2}$A

  4. iN/A

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Correct Option: A
Explanation:

Magnetic dipole moment of small coil of N turns having an area A and a current $i$ is
$M = NiA$

When a current carrying coil is placed in a uniform magnetic field of induction $B$, then a torque $\tau $ acts on it. If $I$ is the current, $n$ is the number of turns and $A$ is the face area of the coil and the normal to the coil makes an angle $\theta $ with $B$, Then

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. $\tau =BInA$

  2. $\tau =B I n A \sin\theta $

  3. $\tau =B I n A \cos\theta $

  4. $\tau =B I n A \tan\theta $

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Correct Option: B
Explanation:

$\vec{\tau} = \vec{M}\times \vec{B}$
$\vec{\tau} = MB \sin\theta$
$= niAB \sin\theta$          $(\because M = niA)$

A rectangular loop carrying a current $i$ is placed in a uniform magnetic field $B$. The area enclosed by the loop is $A$. If there are $n$ turns in the loop, the torque acting on the loop is given by

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. $ni(\bar{A}\times \bar{B})$

  2. $ni(\bar{A}. \bar{B})$

  3. $\dfrac{i(\bar{A}\times \bar{B})}{n}$

  4. $\dfrac{i(\bar{A}. \bar{B})}{n}$

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Correct Option: A
Explanation:

$\vec{\tau} = \vec{M}\times \vec{B}$
$= ni(\vec{A}\times \vec{B})$    $(\because \vec{M} = ni\vec{A})$

When the current through a solenoid increases at a constant rate, the induced current

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. Is a constant and in the direction of inducing current

  2. Is a constant and is opposite to the direction of inducing current

  3. Increases with time and is in the direction of inducing current

  4. Increases with time and is opposite to the direction of inducing current

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Correct Option: D
Explanation:

According to the laws of induction of current, if the source current is increased the induced current is also increased, and the induced current is in the opposite direction to the original current.

If a current is passed in a spring it

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. gets compressed

  2. gets expanded

  3. oscillates

  4. remains unchanged

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Correct Option: A
Explanation:

Spring can be assumed as the coil parallel to each other.
So, when current flows in spring each coil gets current flow in same direction. So, they are attracted to each other which in turn results in compression of spring.

A circular coil of wire is connected to a battery of negligible internal resistance and has magnetic induction $B$ at its centre. If the coil is unwound and rewound to have double the number of turns, and is connected to the same battery, then the magnetic induction at the center 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. $2B$

  2. $4B$

  3. $B$

  4. $0.5B$

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Correct Option: A
Explanation:
Magnetic field, $B$, is directly proportional to number of turns, $n$.  Therefore after rewounding, magnetic field will be double the original value.

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

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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

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

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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}$

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

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