1
MHT CET 2021 22th September Morning Shift
+1
-0

When a battery is connected to the two ends of a diagonal of a square conductor frame of side '$$a$$', the magnitude of magnetic field at the centre will be ( $$\mu_0=$$ permeability of free space)

A
$$\frac{\mu_0}{\sqrt{2} \pi \mathrm{a}}$$
B
$$\frac{\sqrt{2} \mu_0}{\pi \mathrm{a}}$$
C
$$\frac{\mu_0}{\pi \mathrm{a}}$$
D
zero
2
MHT CET 2021 22th September Morning Shift
+1
-0

Two concentric coplanar circular loops of radii '$$r{ }_1$$' and '$$r_2$$' respectively carry currents '$$i_1$$' and '$$\mathrm{i}_2$$' in opposite directions (one clockwise and other anticlockwise). The magnetic induction at the centre of the loops is half that due to '$$i_1$$' alone at the centre. If $$r_2=2 r_1$$, the value of $$\frac{i_2}{i_1}$$

A
$$\frac{1}{4}$$
B
1
C
2
D
$$\frac{1}{2}$$
3
MHT CET 2021 22th September Morning Shift
+1
-0

Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is [$$\mu_0=$$ permeability of free space, $$\mathrm{m}=$$ mass of electron, $$\epsilon_0=$$ permittivity of free space, $$\mathrm{h}=$$ Planck's constant ]

A
$$\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \in_0^3 \mathrm{~h}^5}$$
B
$$\frac{\mu_0 \mathrm{e}^5 \pi \mathrm{m}^3}{8 \epsilon_0^3 \mathrm{~h}^5}$$
C
$$\frac{\mu_0 \mathrm{e}^5 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^2 \mathrm{~h}^4}$$
D
$$\frac{\mu_0 \mathrm{e}^7 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$$
4
MHT CET 2021 21th September Evening Shift
+1
-0

A, B and C are three parallel conductors of equal lengths carrying currents $$\mathrm{I}, \mathrm{I}$$ and $$2 \mathrm{I}$$ respectively. Distance between A and B is '$$x$$' and that between B and C is also '$$x$$'. $$F_1$$ is the force exerted by conductor $$\mathrm{B}$$ on $$\mathrm{A}$$. $$\mathrm{F}_2$$ is the force exerted by conductor $$\mathrm{C}$$ on $$\mathrm{A}$$. Current $$\mathrm{I}$$ in $$\mathrm{A}$$ and $$\mathrm{I}$$ in $$\mathrm{B}$$ are in same direction and current $$2 \mathrm{I}$$ in $$\mathrm{C}$$ is in opposite direction. Then

A
$$\mathrm{F}_1=\mathrm{F}_2$$
B
$$\mathrm{F}_2=2 \mathrm{R}_1$$
C
$$F_1=2 R_2$$
D
$$F_1=-F_2$$
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