JEE Main 2021 (Online) 1st September Evening Shift
MCQ (Single Correct Answer)
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 ____________.
JEE Main 2021 (Online) 1st September Evening Shift
MCQ (Single Correct Answer)
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)
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
MCQ (Single Correct Answer)
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
MCQ (Single Correct Answer)
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