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Solenoid MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

Solenoid is considered one of the most asked concept.
15 Questions around this concept.

Solve by difficulty

Two coaxial solenoids of different radii carry current I in the same direction. Let $\underset{F_{1} }{\rightarrow}$ be the magnetic force on the inner solenoid due to the outer one and $\underset{F_{2} }{\rightarrow}$ be the magnetic force on the outer solenoid due to the inner one. Then :

$\underset{F_{1} }{\rightarrow}$ = $\underset{F_{2} }{\rightarrow}$ = 0

$\underset{F_{1} }{\rightarrow}$ is radially inwards and $\underset{F_{2} }{\rightarrow}$ radially outwards

$\underset{F_{1} }{\rightarrow}$ is radially inwards and $\underset{F_{2} }{\rightarrow}$ = 0

$\underset{F_{1} }{\rightarrow}$ is radially outwards $\underset{F_{2} }{\rightarrow}$ = 0

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A long solenoid has 200 turns per cm and carries a current $i$ . The magnetic field at its center is 6.28 x 10^-2 weber/m². Another long solenoid has 100 turns per cm and it carries a current $i$ /3. The value of the magnetic field at its center is

$1.05\: \times 10^{-4}\; Wb/m^{2}$

$1.05\: \times 10^{-2}\; Wb/m^{2}$

$1.05\: \times 10^{-5}\; Wb/m^{2}$

$1.05\: \times 10^{-3}\; Wb/m^{2}$

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Concepts Covered - 1

Solenoid

Solenoid -

A solenoid is defined as a cylindrical coil of many tightly wound turns of insulated wire with general diameter of the coil smaller than its length.
The solenoid has two ends and one end behaves like the north pole while the opposite end behaves like the south pole. As the length of the solenoid increases, the interior field becomes more uniform and the external field becomes weaker which can be seen from the diagram.

As the current flows a magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of solenoid. Here we will discuss two cases, one with solenoid having finite length and the other when the solenoid is of infinite length.

(i) Finite length solenoid :

Let $\mathrm{n}=$ number of turns per unit length $\frac{N}{L}$
whrere, $\mathrm{N}=$ total number of turns,

$$
l=\text { length of the solenoid }
$$

Magnetic field inside the solenoid at point $P$ is given by -

$$
B=\frac{\mu_0}{4 \pi}(2 \pi n i)[\sin \alpha+\sin \beta]
$$

(ii) Infinite length solenoid -

If the solenoid is of infinite length and the point is well inside the solenoid. So in this case the angle $\alpha$ and $\beta$ will be $\frac{\pi}{2}$. SO if we put this value in the equation of finite length you will get -

$$
B_{i n}=\mu_0 n i
$$

Here again, $\mathrm{n}=$ number of turns per unit length.

Note - Magnetic field outside the solenoid is zero.