B_{due 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)

B_{due 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)

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.

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)

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, R_eq = 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 / R_eq

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.