 ### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2021 (Online) 1st September Evening Shift

There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ____________. A
$${{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$
B
$${{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]$$
C
$${{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} - (x + y)} \right]$$
D
$${{{\mu _0}Ixy} \over {4\pi }}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$

## Explanation Bdue to wire (1) = $${{{\mu _0}I} \over {4\pi y}}\left[ {\sin 90 + \sin {\theta _1}} \right]$$

$$= {{{\mu _0}} \over {4\pi }}{I \over y}\left( {1 + {x \over {\sqrt {{x^2} + {y^2}} }}} \right)$$ ....(1)

Bdue to wire (2) = $${{{\mu _0}} \over {4\pi }}{1 \over x}(\sin 90^\circ + \sin {\theta _2})$$

$$= {{{\mu _0}} \over {4\pi }}{1 \over x}\left( {1 + {y \over {\sqrt {{x^2} + {y^2}} }}} \right)$$ ....(2)

Total magnetic field

$$B = {B_1} + {B_2}$$

$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{1 \over y} + {x \over {y\sqrt {{x^2} + {y^2}} }} + {1 \over x} + {y \over {x\sqrt {{x^2} + {y^2}} }}} \right]$$

$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{{x^2} + {y^2}} \over {xy\sqrt {{x^2} + {y^2}} }}} \right]$$

$$B = {{{\mu _0}I} \over {4\pi }}\left[ {{{x + y} \over {xy}} + {{\sqrt {{x^2} + {y^2}} } \over {xy}}} \right]$$

$$B = {{{\mu _0}I} \over {4\pi xy}}\left[ {\sqrt {{x^2} + {y^2}} + (x + y)} \right]$$

Option (1)
2

### JEE Main 2021 (Online) 1st September Evening Shift

Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by
E = 20cos(2 $$\times$$ 1010 t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\mu$$r = 1)
A
9
B
2
C
$${1 \over 3}$$
D
3

## Explanation

Given, electric field,

E = 20 cos(2 $$\times$$ 1010t $$-$$ 200 x) V/m

Comparing with the standard equation,

E = E0 cos($$\omega$$t $$-$$ kx) V/m, we get

Wave constant, k = 200

Angular frequency, $$\omega$$ = 2 $$\times$$ 1010 rad/s

Speed of the wave, $$v = {\omega \over k} = {{2 \times {{10}^{10}}} \over {200}} = {10^8}$$ m/s

Refractive index, $$\mu = {c \over v} = {{3 \times {{10}^8}} \over {{{10}^8}}} = 3$$

As we know the relation between the refractive index and dielectric constant,

$$\mu = \sqrt {{\varepsilon _r}{\mu _r}}$$

Substituting the value in the above equations, we get

$$3 = \sqrt {{\varepsilon _r}(1)}$$

$${\varepsilon _r} = 9$$

Thus, the dielectric constant of the medium is 9.
3

### JEE Main 2021 (Online) 1st September Evening Shift

Following plots show magnetization (M) vs magnetizing field (H) and magnetic susceptibility ($$\chi$$) vs temperature (T) graph : Which of the following combination will be represented by a diamagnetic material?
A
(a), (c)
B
(a), (d)
C
(b), (d)
D
(b), (c)

## Explanation

For diamagnetic material

$$\chi$$ is independent of temperature and negative.

Magnetisation (M) is directly proportional to H (The relation between magnetisation and magnetising field, M = –mH)

Option (a)

4

### JEE Main 2021 (Online) 1st September Evening Shift

A square loop of side 20 cm and resistance 1$$\Omega$$ is moved towards right with a constant speed v0. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4$$\Omega$$. What should be the value of v0 so that a steady current of 2 mA flows in the loop? A
1 m/s
B
1 cm/s
C
102 m/s
D
10$$-$$2 cm/s

## Explanation

According to given circuit diagram, equivalent resistance between point P and Q.

$${R_{PQ}} = (4 + 4)||(4 + 4)$$

$$= {{8 \times 8} \over {8 + 8}} = 4\,\Omega$$

The equivalent circuit can be drawn as, Equivalent resistance, Req = 4 + 1 = 5 $$\Omega$$

Magnetic field, B = 5T

The side of the square loop, I = 20 cm = 0.20 m

The steady value of the current, I = 2 mA = 2 $$\times$$ 10-3 A

Induced emf, e = Bv0I

Induced current, I = e / Req

Substituting the values in the above equation, we get

$$2 \times {10^{ - 3}} = {{5 \times {v_0} \times 0.2} \over 5}$$

$$\Rightarrow$$ v0 = 10-2 m/s = 1 cm/s

$$\therefore$$ The value of v0 = 1 cm/s, so that a steady current of 2 mA flows in the loop.

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