MHT CET 2026 16th April Morning Shift | Electromagnetic Induction Question 24 | Physics

A metal rod of length 'L' completes the circuit as shown. The area of the circuit is perpendicular to magnetic field 'B'. Total resistance of the circuit is 'R'. The force needed to move the rod in the direction as shown with constant speed 'V' is
MHT CET 2026 16th April Morning Shift Physics - Electromagnetic Induction Question 6 English

MHT CET 2026 16th April Morning Shift Physics - Electromagnetic Induction Question 6 English

$\dfrac{\text{BVL}}{\text{R}}$

$\dfrac{\text{B}^2\text{L}^2\text{V}}{\text{R}}$

$\dfrac{\text{B}^2\text{L}^2\text{V}^2}{\text{R}}$

$\dfrac{\text{BLV}^2}{\text{R}}$

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

Two concentric circular coils having radii $r_1$ and $r_2$ ($r_2 << r_1$) are placed co-axially with centres coinciding. The mutual inductance of the arrangement is (Both coils have single turn, $\mu_0$ = permeability of free space)

$\dfrac{\mu_0 \pi r_2}{2r_1}$

$\dfrac{\mu_0 \pi r_1^2}{2r_2}$

$\dfrac{\mu_0 \pi r_1}{2r_2}$

$\dfrac{\mu_0 r_2^2 \pi}{2r_1}$

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

Two solenoids A and B of equal number of turns have their lengths and radii in the same ratio $1 : 3$. The ratio of the self-inductance of solenoid A to that of B will be

$1 : 1$

$1 : 3$

$1 : 9$

$3 : 1$

MHT CET 2026 15th April Morning Shift

MCQ (Single Correct Answer)

-0

A coil of 'n' turns and resistance $R\,\Omega$ is connected in series with a resistance $R/4$. The combination is moved for time 't' second through flux $\Phi_1$ to $\Phi_2$. The induced current in the circuit is

$\dfrac{4n(\Phi_1 - \Phi_2)}{5RT}$

$\dfrac{5n^2(\Phi_1 - \Phi_2)}{4RT}$

$\dfrac{2n(\Phi_1 - \Phi_2)}{3RT}$

$\dfrac{3n(\Phi_1 - \Phi_2)}{4RT}$

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