MHT CET 2021 20th September Morning Shift | Moving Charges and Magnetism Question 94 | Physics

The magnetic field intensity 'H' at the centre of a long solenoid having 'n' turns per unit length and carrying a current 'I', when no material is kept in it, is

$$\mathrm{\frac{1}{n}}$$

$$\mathrm{\frac{n}{1}}$$

$$\mathrm{nI}$$

$$\mathrm{n^2I}$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

An electron (e) moves in circular orbit of radius 'r' with uniform speed 'V'. It produces magnetic field 'B' at the centre of circle. The magnetic field B is $$\left(\mu_0=\right.$$ permeability of free space)

$$\frac{\mu_0 \mathrm{e}}{4 \pi}\left(\frac{\mathrm{V}}{\mathrm{r}^2}\right)$$

$$\frac{\mu_0 \mathrm{e}}{4 \pi} \mathrm{Vr}^2$$

$$\frac{\mu_0 \mathrm{e}}{4 \pi}\left(\frac{\mathrm{V}}{\mathrm{r}}\right)$$

$$\frac{\mu_0 \mathrm{e}}{4 \pi} \mathrm{Vr}$$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

An electron moves in a circular orbit with uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is [ $\mu_0=$ permeability of free space, $e=$ electronic charge]

$\left(\frac{\mu_0 e v}{4 \pi B}\right)^{1 / 2}$

$\frac{\mu_0 e B}{4 \pi v}$

$\frac{\mu_0 e V}{4 \pi B}$

$\left(\frac{\mu_0 e v}{B}\right)^{1 / 2}$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

A circular coil of radius $$R$$ is carrying a current $$I_1$$ in anti-clockwise sense. A long straight wire is carrying current $$I_2$$ in the negative direction of $$X$$-axis. Both are placed in the same plane and the distance between centre of coil and straight wire is $$d$$. The magnetic field at the centre of coil will be zero for the value of $$d$$ equal to

$$\frac{R}{\pi}\left(\frac{l_2}{l_1}\right)$$

$$\frac{\pi}{R}\left(\frac{l_2}{l_1}\right)$$

$$\frac{\pi}{R}\left(\frac{l_1}{l_2}\right)$$

$$\frac{R}{\pi}\left(\frac{l_1}{l_2}\right)$$

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