Tags Physics Moving Charges and Magnetism

Physics Moving Charges and Magnetism

Get insights from 109 questions on Physics Moving Charges and Magnetism, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Moving Charges and Magnetism

Follow Ask Question

109

Questions

Discussions

Active Users

Followers

New answer posted

a month 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

0 Follower 1 View

Raj Pandey

Contributor-Level 9

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

0 0

New answer posted

a month 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?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

N = qvB

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

$μ q B S = m v$

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

0 0

New answer posted

a month 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-

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Apply Ampere's Law

For inside

$B 2 π r = μ_{0} J π r^{2}$

$\Rightarrow B \propto r$

For outside

$B 2 π r = μ_{0} J π R^{2}$

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

0 0

New answer posted

a month 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:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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

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

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

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

0 0

New answer posted

a month 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:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

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

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

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

0 0

New answer posted

a month 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.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

a month 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.

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

d = Rsin 60°

⇒ d = (mv√3)/ (qB 2) = (√3v)/ (2aB)

0 0

New answer posted

a month 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:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

a month 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:

0 Follower 2 Views