MHT CET 2023 13th May Evening Shift | Moving Charges and Magnetism Question 45 | Physics

The ratio of magnetic field at the centre of the current carrying circular loop and magnetic moment is $$X$$. When both the current and radius are doubled, then the ratio will be

$$2 X$$

$$\frac{X}{2}$$

$$\frac{X}{4}$$

$$\frac{X}{8}$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

-0

A circular current carrying coil has radius $$R$$. The magnetic induction at the centre of the coil is $$B_C$$. The magnetic induction of the coil at a distance $$\sqrt{3} R$$ from the centre along the axis is $$B_A$$. The ratio $$B_A: B_C$$ is

$$1: 3$$

$$1: 8$$

$$8: 1$$

$$27: 1$$

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

-0

A circular coil of radius '$$r$$' and number of turns ' $n$ ' carries a current '$$I$$'. The magnetic fields at a small distance '$$h$$' along the axis of the coil $$\left(B_a\right)$$ and at the centre of the coil $$\left(\mathrm{B}_{\mathrm{c}}\right)$$ are measured. The relation between $$B_c$$ and $$B_a$$ is

$$\mathrm{B_c=B_a\left(1+\frac{h^2}{r^2}\right)}$$

$$\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{1}{2}}$$

$$\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{\frac{3}{2}}$$

$$\mathrm{B}_{\mathrm{c}}=\mathrm{B}_{\mathrm{a}}\left(1+\frac{\mathrm{h}^2}{\mathrm{r}^2}\right)^{-\frac{3}{2}}$$

MHT CET 2023 13th May Morning Shift

MCQ (Single Correct Answer)

-0

Two concentric circular coils A and B have radii $$20 \mathrm{~cm}$$ and $$10 \mathrm{~cm}$$ respectively lie in the same plane. The current in coil A is $$0.5 \mathrm{~A}$$ in anticlockwise direction. The current in coil B so that net field at the common centre is zero, is

$$0.5 \mathrm{~A}$$ in anticlockwise direction

$$0.25 \mathrm{~A}$$ in anticlockwise direction.

$$0.25 \mathrm{~A}$$ in clockwise direction.

$$0.125 \mathrm{~A}$$ in clockwise direction.

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