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Physics Moving Charges and Magnetism

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3 months ago

Physics Moving Charges and Magnetism Class 12th

If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by proton in a cyclotron to achieve one sixth of the speed of light is_________.

$[m_{p} = 1.67 * 1 0^{- 27} k g, e = 1.6 * 1 0^{- 19} C, S p e e d o f l i g h t = 3 * 1 0^{8} m ? / ? s]$

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alok kumar singh

Contributor-Level 10

2nqV = $\frac{1}{2} m v^{2}$

$n = \frac{m v^{2}}{4 q V}$

$= \frac{1.67 * 1 0^{- 27} * {(0.5 * 1 0^{8})}^{2}}{4 * 1.6 * 1 0^{- 19} * 12 * 1 0^{3}} = 543$

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Physics Moving Charges and Magnetism Class 12th

The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is: (Take r < a)

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Vishal Baghel

Contributor-Level 10

$B_{a x i s} = \frac{μ_{0} N l a^{2}}{2 {(a^{2} + r^{2})}^{3 / 2}}$

$B_{c e n t} = \frac{μ_{0} N I}{2 a}$

Fractional change = $\frac{B_{c e n t r e} - B_{a x i s}}{B_{c e n t r e}}$

$= 1 - \frac{B_{a x i s}}{B_{c e n t r e}}$

$= 1 - (1 - \frac{3 r^{2}}{2 a^{2}}) [r < < a]$

$= \frac{3 r^{2}}{2 a^{2}}$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

A 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:

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Vishal Baghel

Contributor-Level 10

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

1 unit Area $\frac{l}{π R^{2}}$ $\frac{}{^{}}$

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

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

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

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

$B = \frac{μ_{0}}{2 π} \frac{l}{R^{2}} r$

$B \propto r$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

A wire X of length 50cm carrying a current of 2A is placed parallel to a long wire Y of length 5m. The wire Y current of 3A. The distance between two wires is 5cm and currents flow in the same direction. The force acting on the wire Y is

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Payal Gupta

Contributor-Level 10

F_x-y = $\frac{μ_{0} i_{1} i_{2}}{2 π r} * (. 5)$

$= \frac{2 * 1 0^{- 7} * 6 * 0.5}{0.05}$

$= \frac{6}{5} * 1 0^{- 5} N = 1.2 * 1 0^{- 5}$

Force on Y = 1.2 * 105N towards 'x'

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3 months ago

Physics Moving Charges and Magnetism Class 12th

A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$ . If an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle, its speed will double after a time:

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alok kumar singh

Contributor-Level 10

Since magnetic force cannot change the speed. So only electric field which is along -direction will change the speed along -direction only.

$v_{x} = \frac{E_{0} q}{m} t$ but $v_{y} = v_{0}$

$2 v_{0} = \sqrt[]{v_{x}^{2} + v_{y}^{2}}$

4 v_{0}^{2} = \frac{E_{o}^{2} q^{2} t^{2}}{m^{2}} + v_{0}^{2}

t = \frac{\sqrt[]{3} m v_{o}}{q E_{o}}

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3 months ago

Physics Moving Charges and Magnetism Class 12th

Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. It the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be:

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Vishal Baghel

Contributor-Level 10

Radius of circular path R = $\sqrt[]{\frac{2 m k}{q B}}$ $\frac{}{}$

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

$\frac{q_{1}}{q_{2}} = \sqrt[]{\frac{m_{1}}{m_{2}}} * \frac{R_{2}}{R_{1}} = \sqrt[]{\frac{9}{4}} * \frac{5}{6} = \frac{5}{4}$

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Physics Moving Charges and Magnetism Class 12th

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3s and 4s respectively. If their moments of inertia are in the ratio of 3 : 2, then the ratio of their magnetic moments will be:

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alok kumar singh

Contributor-Level 10

T₁ = 3 sec T₂ = 4sec

$T = 2 π \sqrt[]{\frac{I}{μ B}}$

$\frac{I_{1}}{I_{2}} = \frac{3}{2} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt[]{\frac{l_{1} μ_{2}}{μ_{1} l_{2}}}$

$\Rightarrow {(\frac{3}{4})}^{2} = \frac{I_{1}}{I_{2}} (\frac{μ_{2}}{μ_{1}}) \Rightarrow \frac{μ_{2}}{μ_{1}} = \frac{l_{2}}{l_{1}} * \frac{9}{16} = \frac{2}{3} * \frac{9}{16} = \frac{3}{8}$ $\Rightarrow \frac{μ_{1}}{μ_{2}} = \frac{8}{3}$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 * 10^-4 Wbm^-2. The frequency of revolution of the electron will be

(Take mass of electron = 9.0 * 10^-31kg)

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Payal Gupta

Contributor-Level 10

f = $\frac{q B}{2 π m} = \frac{1.6 * 1 0^{- 19} * 1.0 * 1 0^{- 4}}{2 * 3.14 * 9 * 1 0^{- 31}} = 0.028 * 1 0^{8} H z = 2.8 * 1 0^{6} H z$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

The electric current in a circular coil of 2 turns produces a magnetic induction B₁ at its centre. The coil is unwound and is rewound into circular coil of 5 turns and at its centre. The coil is unwound and is rewound into a circular coil of 5 turns and the same current produces a magnetic induction B₂ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is

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Payal Gupta

Contributor-Level 10

$B_{1} = \frac{N_{1} μ_{0} l}{2 R}$ given N1 = 2

When new loop is made the length of wire remains same, so

$N_{2} * 2 π r = N_{1} * 2 π R \Rightarrow r = \frac{N_{1} R}{N_{2}}$

$\Rightarrow \frac{B_{2}}{B_{1}} = {(\frac{N_{2}}{N_{1}})}^{2} = \frac{25}{4}$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

As shown in the figure, a metallic rod of linear density 0.45 km^-¹ is lying horizontally on a smooth incline plane which makes an angle of 45° 5with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be:

{Use g = m/s²}

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Vishal Baghel

Contributor-Level 10

Linear density, $λ =$ 0.45 kg/m

Let length = $l$ $∴$ = m = 0.45 $l$

B = 0.15 T

For equilibrium of rod :-

mg sin 45° = FB sin 45°

$\Rightarrow (0.45 l) g = l l B$

So, $l = \frac{0.45 * 10}{0.15} = 30 A$ $\frac{}{}$

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