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1

### JEE Main 2021 (Online) 27th August Evening Shift

A constant magnetic field of 1T is applied in the x > 0 region. A metallic circular ring of radius 1m is moving with a constant velocity of 1 m/s along the x-axis. At t = 0s, the centre of O of the ring is at x = $$-$$1m. What will be the value of the induced emf in the ring at t = 1s? (Assume the velocity of the ring does not change.) A
1V
B
2$$\pi$$V
C
2V
D
0V

## Explanation

emf = BLV

= 1 . (2R) . 1

= 2V
2

### JEE Main 2021 (Online) 27th August Morning Shift

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
A
lighter ion will be deflected less than heavier ion
B
lighter ion will be deflected more than heavier ion
C
both ions will be deflected equally
D
no ion will be deflected

## Explanation

$$r = {P \over {qB}} = {{\sqrt {2mk} } \over {qB}}$$

Given they have same kinetic energy

$$r \propto {{\sqrt m } \over q}$$

$${{{r_1}} \over {{r_2}}} = {{\sqrt 4 } \over 2} \times {3 \over {\sqrt {16} }} = {3 \over 4}$$

$${r_2} = {{4{r_1}} \over 3}$$ (r2 is for heavier ion and and r1 is for lighter ion) $$\sin \theta = {d \over R}$$

$$\theta$$ $$\to$$ Deflection

$$\theta \propto {1 \over R}$$

(R $$\to$$ Radius of path)

$$\because$$ R2 > R1 $$\Rightarrow$$ $$\theta$$2 < $$\theta$$1
3

### JEE Main 2021 (Online) 27th August Morning Shift

A bar magnet is passing through a conducting loop of radius R with velocity $$\upsilon$$. The radius of the bar magnet is such that it just passes through the loop. The induced e.m.f. in the loop can be represented by the approximate curve : A B C D ## Explanation $$\to$$ When magnet passes through centre region of solenoid, no current / Emf is induced in loop.

$$\to$$ While entering flux increases so negative induced emf

$$\to$$ While leaving flux decreases so positive induced emf.
4

### JEE Main 2021 (Online) 26th August Evening Shift

A light beam is described by $$E = 800\sin \omega \left( {t - {x \over c}} \right)$$. An electron is allowed to move normal to the propagation of light beam with a speed of 3 $$\times$$ 107 ms$$-$$1. What is the maximum magnetic force exerted on the electron?
A
1.28 $$\times$$ 10$$-$$18 N
B
1.28 $$\times$$ 10$$-$$21 N
C
12.8 $$\times$$ 10$$-$$17 N
D
12.8 $$\times$$ 10$$-$$18 N

## Explanation

$${{{E_0}} \over C} = {B_0}$$

$${F_{\max }} = e{B_0}V$$

$$= 1.6 \times {10^{ - 19}} \times {{800} \over {3 \times {{10}^8}}} \times 3 \times {10^7}$$

$$= 12.8 \times {10^{ - 18}}$$ N

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