Physics Moving Charges and Magnetism

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the centre of the ring is

1 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

When the ring rotates about its axis with a uniform frequency fHz, the current flowing in the ring is

I=q/T=qf

Magnetic field at the centre of the ring is

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

In gravity free space, a bead of charge $1 μ C$ and mass 3mg is threaded on a rough rod of friction coefficient $μ$ = 0.3. A magnetic field of magnitude 0.2 T exists perpendicular to the rod. The bead is projected along the rod with a speed 4m/s. How much distance (in m) will the bead cover before coming to rest?

46 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

N = qvB

$- μ q v B = \frac{m d v}{d t} = m v \frac{d v}{d S}$

$μ q B S = m v$

$S = \frac{3 \times 1 0^{- 6} \times 4}{0.3 \times 1 0^{- 6} \times 0.2} = 200 m$

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

An infinite cylindrical current carrying wire carries a current that is uniformly distributed over its cross section. The magnetic field as a function of distance form it’s axis is-

76 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Apply Ampere’s Law

For inside

$B 2 π r = μ_{0} J π r^{2}$

$\Rightarrow B \propto r$

For outside

$B 2 π r = μ_{0} J π R^{2}$

$\Rightarrow B \propto \frac{1}{r}$

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New answer posted•a year ago

Physics Moving Charges and Magnetism 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

Physics Moving Charges and Magnetism 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

Physics Moving Charges and Magnetism Class 12th

An electron moving with a velocity of 2 m/s along +ve x-axis at a point in a magnetic field experiences a force of 2N along -ve y-axis. If the electron moves with of a velocity of 2 m/s along +ve y-axis at the same point, it experiences a force of 2N along +ve x-axis. Find force that the electron would experience if it were moving with a velocity of 2m/s along z-axis.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

v? = 2î F? = -2?
v? = 2? F? = -2î
⇒ B? is along -k? Hence v? = 2k? ⇒ F? = 0

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Magnetic field exists in the region between lines PQ and RS as shown in the figure. A particle having charge to mass ratio of a enters the magnetic field as shown in the figure with a speed v. Lines PQ and RS are parallel to each other. Find the value of d such that the charged particle exits the magnetic field region with velocity perpendicular to line RS.

17 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

d = Rsin 60°

⇒ d = (mv√3)/ (qB 2) = (√3v)/ (2aB)

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

A proton of mass m and charge q enters a region of uniform magnetic field of a magnitude B with a velocity v directed perpendicular to the magnetic field. It moves in a circular path and leaves the magnetic field after completing a quarter of a circle. The time spent by the proton inside the magnetic field is proportional to:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The time spent in magnetic field (t) is independent of velocity of charge.
t = (1/4) (2πm/qB)

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

AB µ r² long straight wire with a circular cross-section having radius R, is carrying a steady current I: The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r(r < R) from its centre will be:

4 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

for $π R^{2} A r e a \to' l'$ current\

1 unit Area ® $\frac{l}{π R^{2}}$

$π r^{2} \to \frac{l}{π R^{2}} \times π r^{2}$

$i = \frac{l r^{2}}{R^{2}}$

Now, consider Amperian loop of radius small ‘r’ ln Amperian loop magnetic field will be tangential to the amperian loop.

$? \vec{B} . d \vec{i} = μ_{0} l_{e n c l o s e d}$

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New answer posted•a year ago

Physics Moving Charges and Magnetism Force between Two Parallel Currents

What is the formula for the force per unit length between two parallel current-carrying conductors?

1 Views|Posted a year ago

Asked by Anushree

Answered by

Chandra Pruthi | Beginner-Level 5

a year ago

The exact formula for the magnetic force per unit length between two parallel current-carrying conductors:

$\frac{F}{l} = \frac{?_{0} I 1 I 2}{2 ? d}$ For more information, check: Force between parallel conductors

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Physics Moving Charges and Magnetism

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