JEE Main Electromagnetic Induction PYQ

6. (JEE Main 2023 (Online) 24th January Evening Shift)

A metallic rod of length 'L' is rotated with an angular speed of ' $ω$ ' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :

JEE Main 2023 (Online) 24th January Evening Shift Physics - Electromagnetic Induction Question 33 English

A. $\frac{1}{2} B^{2} L^{2} ω$

B. $\frac{1}{2} {BL}^{2} ω$

C. $\frac{1}{4} {BL}^{2} ω$

D. $\frac{1}{4} B^{2} L ω$

Correct Option is (B)

Velocity of centre of $rod v = \frac{ω L}{2}$

So, $emf = B \cdot v L$ $= \frac{B ω L^{2}}{2}$

JEE Main 2023 (Online) 24th January Evening Shift Physics - Electromagnetic Induction Question 33 English Explanation

7. (JEE Main 2023 (Online) 15th April Morning Shift)

A $20 cm$ long metallic rod is rotated with $210 rpm$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field $0.2 T$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is ____________ $mV$ .

Take $π = \frac{22}{7}$

Correct answer is 88

Given that the rod is rotating at 210 rpm, we first convert this to radians per second:

$ω = 210 \cdot \frac{2 π rad}{60 s} = 22 rad / s$

Now, we can find the linear velocity $v$ of the tip of the rod:

$v = ω r$

where $r$ is the length of the rod (0.2 m).

$v = 22 rad / s \cdot 0.2 m = 4.4 m / s$

Now, we can find the emf developed between the center and the ring using the formula:

$ϵ = \frac{1}{2} B ℓ v$

where $B$ is the magnetic field (0.2 T), $ℓ$ is the length of the rod (0.2 m), and $v$ is the linear velocity (4.4 m/s).

$ϵ = \frac{1}{2} \cdot 0.2 T \cdot 0.2 m \cdot 4.4 m / s = 0.088 V$

To express this value in mV, we can simply multiply it by 1000:

$ϵ = 0.088 V \cdot 1000 = 88 mV$

So the emf developed between the center and the ring is 88 mV.

8. (JEE Main 2023 (Online) 11th April Evening Shift)

A metallic cube of side $15 cm$ moving along $y$ -axis at a uniform velocity of $2 {ms}^{- 1}$ . In a region of uniform magnetic field of magnitude $0.5 T$ directed along $z$ -axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV.

JEE Main 2023 (Online) 11th April Evening Shift Physics - Electromagnetic Induction Question 27 English

Correct answer is 150

$\begin{aligned} q E = e V B \\ E = V B \end{aligned}$

$\begin{aligned} Δ V & = (E . d) \\ = (V B) \times 0.15 \\ = 2 \times \frac{1}{2} \times 0.15 V \\ = 0.15 V = 150 mV \end{aligned}$

9. (JEE Main 2022 (Online) 26th June Evening Shift)

A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s^{$-$ 1}. If the horizontal component of earth's magnetic field is 0.2 $\times$ 10^{$-$ 4} T, then emf induced between the two ends of the conductor is :

A. 5 $μ$ V

B. 50 $μ$ V

C. 5 mV

D. 50 mv

Correct Option is (B)

$E m f = \frac{1}{2} B ω l^{2}$

$= \frac{1}{2} \times 0.2 \times 10^{- 4} \times 5 \times 1^{2}$ V

= 0.5 $\times$ 10^{$-$ 4} V

= 50 $μ$ V

10. (JEE Main 2022 (Online) 27th June Evening Shift)

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 $\times$ 10^{$-$ 3} T and the angle of dip is 45 $^{\circ}$ . The emf induced in the rod is ___________ mV.

Correct answer is 16

$E = B l v$

$= 4 \times 10^{- 3} \times \frac{20}{100} \times 20$ Volts

$= 16$ mV