Motional EMF - NEET Physics Chapterwise MCQs & PYQs

NEET Motional EMF MCQs & PYQs

Question 11:

moderate

A wheel with ten metallic spockes each 0.50 m long is rotated with a speed of 120 rev/min in a plane normal to the earth’s magnetic field at the place. If the magnitude of the field is 0.4 Gauss, the induced e.m.f. between the axle and the rim of the wheel is equal to :

 

Using the formula for induced e.m.f. in a rotating conductor:

$$e = \frac{1}{2} B \omega L^2 = \frac{1}{2} B (2\pi f) L^2$$

Substituting

$$e = \frac{1}{2} \times (0.4 \times 10^{-4}) \times (4\pi) \times (0.5)^2 = 6.28 \times 10^{-5}\text{ V}$$

Question 12:

easy

An aeroplane in which the distance between the tips of the wings is \(50\text{ m}\) is flying horizontally with a speed of \(360\text{ km/hr}\) over a place where the vertical component of earth’s magnetic field is \(2 \times 10^{-4}\text{ Wbm}^{-2}\). The potential difference between the tips of the wings would be:

Induced EMF is \(e = B_v l v\). Converting speed: \(v = 360\text{ km/h} = 100\text{ m/s}\). Thus, \(e = (2 \times 10^{-4}) \times 50 \times 100 = 1.0\text{ V}\).

Question 13:

easy

Assertion (A): The probability of burn out of a dc motor is maximum, when the motor is just switched on.


Reason (R): No back emf is developed in the armature of dc motor, when it is just switched on.


 

Assertion (A) is true: When a DC motor starts, its speed is zero, thus the back EMF (\(epsilon_b\)) is zero. This leads to the maximum current (\(I = \frac{V - \epsilon_b}{R_a}\)) drawn from the supply, which can cause burnout. Reason (R) is true: Back EMF is proportional to the motor's angular speed (\(epsilon_b = k\Phi\omega\)), so it is zero at startup (\(omega = 0\)). Reason (R) correctly explains Assertion (A).