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Magnetic field due to a current in a solenoid - class-XI

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Why does toroid have a higher magnetic field than a solenoid?

  1. It is magnetically efficient.

  2. It is a straight core.

  3. It has an open loop core.

  4. It has a closed loop core.


Correct Option: D
Explanation:
Due to the symmetry and being closed loop core.
Totoid has higher magnetic field and thus higher inductor and Q-facator

Which among the following is not an electronic component used in constructing a toroid?

  1. copper wires

  2. laminated iron

  3. ferrite core

  4. aluminium powder


Correct Option: D
Explanation:
Toroids are inductors and transformers which use magnetic cores.
These cores are of ferromagnetic material such as
Laminated iron, iron powder or ferrite.

The leakage flux of a toroid is less because

  1. It is asymmetrical.

  2. It has an open-loop core.

  3. It is symmetrical.

  4. It has a straight core.


Correct Option: C
Explanation:
Toroidal is a surface of revolution with hole in middle.
This is a solid body used for toroid production.
The advantage of this shape is that, due to its symmetry amount of magnetic flux that escapes outside is low.

Which of the following is true for a toroid?

  1. low inductance and Q factor

  2. high inductance and Q factor

  3. high inductance and low Q-factor

  4. low inductance and high Q-factor


Correct Option: B
Explanation:

Due to the symmetry and being closed loop core.
Toroid has higher magnetic field and thus higher inductor and Q-factor

A toroid is                  solenoid

  1. a finite

  2. an endless

  3. straight

  4. either A or B


Correct Option: B
Explanation:

A Toroid is a solenoid which is bent in the form of a ring with its both the ends joined together.  Since both the ends are merged, it appears endless.

A solenoid of length 0.4 m, having 500 turns and 3A current flows through it. A coil of radius 0.01 m and have 10 turns and carries current of 0.4 A has to placed such that its axis is perpendicular to the axis of solenoid ,then torque on coil will be

  1. $5.92\times 10^{-7} N.m$

  2. $5.92\times 10^{-5} N.m$

  3. $5.92\times 10^{-4} N.m$

  4. $5.92\times 10^{-3} N.m$


Correct Option: A
Explanation:
Magnetic field due to larges solenoid
$B=\dfrac {\mu _{0} NI}{l}=\dfrac {4\times 10^{-7}\times 500 \times 3}{0.4}$
$N=500$ 
$I=3$
$l=0.4$
Torque $M$ solenoid $I\times $ Area of corn section $\times B$
$=0.4\times \pi 0.01^{2}\times 4\pi \times \dfrac{10^{-7}\times 500 \times 3}{0.4}$
 orque $=\pi.0.01^{2} \times 4\pi \times 10^{-7}\times 500 \times 3$
Torque $=5.92\times 10^{-7}N-m$

A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury when a current from the battery is started in the coil through the mercury

  1. the wire oscillates

  2. the wire continues making contact

  3. the wire breaks contact just when the current is passed

  4. the mercury will expand by heating due to passage of current


Correct Option: A
Explanation:

According to the Biot Savart's law, a current carrying conductor produces a magnetic field in its surroundings. This will cause the helix to oscillate across a mean position.

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


Correct Option: C
Explanation:

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

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

$= \mu _0 N _0 i = B$


$\Rightarrow B _1 = B$

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. $2.67 \times10^{-4} T$


Correct Option: B
Explanation:

Given $d = 14 cm=0.14 m$    

           $l = 60cm = 0.6 m$

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

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

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

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

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

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. $0.8$


Correct Option: D
Explanation:

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

A circular coil of wire of $n$ turns has a radius $r$ and carries a current $i$. Its magnetic dipole moment is $M$. Now the coil is unwound and again rewound into a circular coil of half the initial radius and the same current is passed through it, then the dipole moment of this new coil is :

  1. $\dfrac{M}{2}$

  2. $\dfrac{M}{4}$

  3. $M$

  4. $2M$


Correct Option: A
Explanation:

The length of remains same $N _1\pi*r=N _2\pi*r/2$


$N _2=2n$

$M=NAI=nI\pi*r^2$

$M _2=N _2I*\pi*(r/2)^2=2n*I\pi*(r^2/4)$

$\dfrac{M _2}{M}=\dfrac{1}{2}$

$M _2=\dfrac{M}{2}$

A rectangular coil of wire of $500$ turns of area $10\times 5cm^{2}$ carries a current of $2 A$ in a magnetic field of induction $2\times 10^{-3}T$ . If the plane of the coil is parallel to the field. The torque on the coil is (in$ Nm$):

  1. $0.1$

  2. $0.01$

  3. $0.001$

  4. $1$


Correct Option: B
Explanation:

The torque on the rectangular coil due to presence of magnetic field is given, 

$\tau=NIAB\sin\theta$
where number of turns $N=500$,
Current in the coil $I=2A$, 
Area of the coil  $A=(10\times 5)10^{-4} m^2$,
Magnetic field $B=2\times 10^{-3}T$,
Angle between area and magnetic field vector is $\theta$ 
As area vector is always normal to plane and given the plane is parallel to field, so the angle between area and field is $90^o$.
So, $\tau=500\times 2\times (10\times 5)\times 10^{-4}\times (2\times 10^{-3})sin90=0.01 \, Nm$

A current I ampere flows along an infinitely long straight thin walled hollow metallic cylinder of radius r . The magnetic field at any point inside the cylinder at a distance x from the axis of the cylinders is :

  1. $\dfrac{\mu _{0}I}{4\pi r}$

  2. $\dfrac{\mu _{0}I}{2\pi r}$

  3. $\dfrac{\mu _{0}I}{2\pi x}$

  4. zero


Correct Option: D
Explanation:

$\oint B\cdot dl=\mu _0 I _{enclosed}$
Here,
$I _{enclosed}=0$
So,  B $=0$

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

  1. i NA

  2. $i^{2}$ NA

  3. i $N^{2}$A

  4. iN/A


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

  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 $


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

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

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

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

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


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

  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


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

  1. gets compressed

  2. gets expanded

  3. oscillates

  4. remains unchanged


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 :

  1. $2B$

  2. $4B$

  3. $B$

  4. $0.5B$


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 $= 1.67 \times 10^{-27} kg$

  1. 2.3 cm

  2. 5.35 cm

  3. 4.35 cm

  4. 6.35 cm


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

  1. 1000

  2. 3000

  3. 5000

  4. 10000


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?

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


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}$

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

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

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