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1

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

A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?
A
$${{{x^2}} \over {{a^2}}}$$
B
$${{{a^2}} \over {{x^2}}}$$
C
$${{{x^2}} \over {{b^2} - {a^2}}}$$
D
$${{{b^2} - {a^2}} \over {{x^2}}}$$

## Explanation

when x < a

$${B_1}(2\pi x) = {\mu _0}\left( {{{{i_0}} \over {\pi {a^2}}}} \right)\pi {x^2}$$

$$B(2\pi x) = {{{\mu _0}{i_0}{x^2}} \over {{a^2}}}$$

$${B_1} = {{{\mu _0}{i_0}x} \over {2\pi {a^2}}}$$ .... (1)

when a < x < b

$${B_2}(2\pi x) = {\mu _0}{i_0}$$

$${B_2} = {{{\mu _0}{i_0}} \over {2\pi x}}$$ ..... (2)

$${{{B_1}} \over {{B_2}}} = {{{\mu _0}{i_0}{x \over {2\pi {a^2}}}} \over {{{{\mu _0}{i_0}} \over {2\pi x}}}} = {{{x^2}} \over {{a^2}}}$$
2

### 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
3

### 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
4

### 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.

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