Motion in Magnetic Field

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New answer posted•a year ago

Motion in Magnetic Field Class 12th

A neutral point is obtained at the Centre of a vertical circular coil carrying current. The angle between the plane of the coil and the magnetic meridian is

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$θ = 90 °$

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New answer posted•a year ago

Motion in Magnetic Field Class 12th

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then:

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$R = \frac{\sqrt[]{2 E m}}{B . q}$

$\Rightarrow R \times \frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{\sqrt[]{4}}{2} \times \frac{3}{\sqrt[]{16}} = \frac{3}{4}$

$\Rightarrow R_{2} = \frac{4 R_{1}}{3}$

$s i n θ = \frac{d}{R} \Rightarrow θ α \frac{1}{2} \Rightarrow θ_{2} < θ_{1}$

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New answer posted•a year ago

Motion in Magnetic Field Class 12th

A cell having N turns is would tightly in the form of a spiral with inner and outer radii ‘a’ and ‘b’ respectively. Find the magnetic field at centre, when a current I passes through coil:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$d N = (\frac{N}{b - a}) d r$

$∴ {(d B)}_{a t c e n t r e} = \frac{μ_{0} l d N}{2 r}$

$B = \frac{μ_{0} l N}{2 (b - a)} l n \frac{b}{a}$

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New answer posted•a year ago

Motion in Magnetic Field Class 12th

A charge particle is moving in uniform magnetic field $(2 \hat{i} + 3 \hat{j})$ T. If it has an acceleration of $(α \hat{i} - 4 \hat{j}) m / s^{2}$ , then the value of will be

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\vec{a} ⊥ \vec{B}$

$\Rightarrow \vec{a} . \vec{B} = 0$

$\Rightarrow (2 α - 12) = 0$

= 6

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Motion in Magnetic Field

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A neutral point is obtained at the Centre of a vertical circular coil carrying current. The angle between the plane of the coil and the magnetic meridian is

Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then:

A cell having N turns is would tightly in the form of a spiral with inner and outer radii ‘a’ and ‘b’ respectively. Find the magnetic field at centre, when a current I passes through coil:

A charge particle is moving in uniform magnetic field (2i^+3j^) T. If it has an acceleration of (αi^−4j^)m/s2 , then the value of will be

A charge particle is moving in uniform magnetic field $(2 \hat{i} + 3 \hat{j})$ T. If it has an acceleration of $(α \hat{i} - 4 \hat{j}) m / s^{2}$ , then the value of will be