Tag: magnetic field on the axis of a toroid

Questions Related to magnetic field on the axis of a toroid

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, $N \times \pi d = 28 m.$

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

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

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

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