6. ⇒ (MHT CET 2023 11th May Evening Shift )

The self inductance ' $L$ ' of a solenoid of length ' $l$ ' and area of cross-section ' $A$ ', with a fixed number of turns ' $N$ ' increases as

A. both $l$ and $A$ increase

B. $l$ decreases and $A$ increases

C. $l$ increases and A decreases

D. both $l$ and A decrease

Correct Option is (B)

The equation for self-inductance is:

$L = \frac{μ_{0} N^{2} A}{l}$

$∴$ The self-inductance $L$ of a solenoid of length $l$ and area of cross section ' $A$ ', with a fixed number of turns ' $N$ ' increases as $l$ decreases and A increases.

7. ⇒ (MHT CET 2023 11th May Morning Shift )

A coil of radius ' $r$ ' is placed on another coil (whose radius is $R$ and current flowing through it is changing) so that their centres coincide $(R ≫ r)$ . If both the coils are coplanar then the mutual inductance between them is ( $μ_{0} =$ permeability of free space)

A. $\frac{μ_{0} π R^{2}}{2 r}$

B. $\frac{μ_{0} π r^{2}}{2 R}$

C. $\frac{μ_{0} π r^{2}}{R}$

D. $μ_{0} π R^{2}$

Correct Option is (B)

Magnetic field, $B = \frac{μ_{0} I}{2 R}$

Flux passing through the coil,

$\begin{aligned} ϕ & = B \times π r^{2} \\ ∴ ϕ & = \frac{μ_{0} I}{2 R} \times π r^{2} \end{aligned}$

Mutual Inductance, $M = \frac{ϕ}{I}$

$∴ M = \frac{\frac{μ_{0} I}{2 R} \times π r^{2}}{I} = \frac{μ_{0} π r^{2}}{2 R}$

8. ⇒ (MHT CET 2023 11th May Morning Shift )

When a current of $1 A$ is passed through a coil of 100 turns, the flux associated with it is $2.5 \times 10^{- 5} Wb /$ turn. The self inductance of the coil in millihenry is

A. 40

B. 25

C. 4

D. 2.5

Correct Option is (D)

Given: $N = 100, I = 1 A, ϕ = 2.5 \times 10^{- 5} Wb /$ turn

Self-inductance of coil,

$\begin{aligned} L & = \frac{N ϕ}{I} \\ ∴ L & = \frac{100 \times 2.5 \times 10^{- 5}}{1} = 2.5 \times 10^{- 3} H = 2.5 mH \end{aligned}$

9. ⇒ (MHT CET 2023 11th May Morning Shift )

The mutual inductance of a pair of coils, each of ' $N$ ' turns, is ' $M$ ' henry. If a current of ' $I$ ' ampere in one of the coils is brought to zero in ' $t$ ' second, the e. m. f. induced per turn in the other coil in volt is

A. $\frac{MI}{t}$

B. $\frac{NMI}{t}$

C. $\frac{NM}{It}$

D. $\frac{MI}{Nt}$

Correct Option is (A)

Emf induced due to mutual inductance per coil,

$e = \frac{M Δ I}{Δ t}$

Here, the current changes from I to zero in time $t$ seconds.

$∴ e = \frac{M I}{t}$

10. ⇒ (MHT CET 2023 11th May Morning Shift )

To manufacture a solenoid of length $1 m$ and inductance $1 mH$ , the length of thin wire required is

(cross - sectional diameter of a solenoid is considerably less than the length)

A. 0.10 m

B. 0.10 km

C. 1 km

D. 10 km

Correct Option is (B)

Inductance of solenoid, $L = \frac{μ_{0} N^{2} A}{l}$ .....(i)

where, $l =$ length of solenoid.

$A = π r^{2} = area of solenoid$

Let ' $x$ ' be length of wire required.

$\begin{aligned} ∴ x & = circumference of solenoid \times no. of turns \\ = 2 π rN . \\ ∴ N & = \frac{x}{2 π r} \end{aligned}$

Substituting in equation (i),

$\begin{aligned} L = \frac{μ_{0} (\frac{x^{2}}{4 π^{2} r^{2}}) \times π r^{2}}{l} \\ ∴ L = \frac{μ_{0} x^{2}}{4 π l} \\ ∴ x^{2} = \frac{4 π L l}{μ_{0}} \\ ∴ x = \sqrt{\frac{4 π L l}{μ_{0}}} \end{aligned}$

Substituting the values,

$\begin{aligned} x = \sqrt{\frac{4 \times π \times 10^{- 3} \times 1}{4 π \times 10^{- 7}}} \\ x = \sqrt{10^{4}} m \\ x = 0.10 km \end{aligned}$

⇒ Electromagnetic Induction