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Physics Moving Charges and Magnetism

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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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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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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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a year 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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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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a year 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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New answer posted

a year ago

Physics Moving Charges and Magnetism Class 12th

The current sensitivity of a galvanometer can be increased by:

(a) decreasing the number of turns

(b) increasing the magnetic field

(c) decreasing the area of the coil

(d) decreasing the torsional constant of the spring

Choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Current sensitivity, $\frac{}{}$ $S i = \frac{N A B}{k}$

$N i A B = k θ$

$θ = (\frac{N A B}{k}) i$

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a year ago

Physics Moving Charges and Magnetism Class 12th

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alok kumar singh

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 * 10^-5N towards 'x'

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New answer posted

a year ago

Physics Moving Charges and Magnetism Class 12th

A wire of length 314cm carrying current of 14A is bent to form a circle. The magnetic moment of the coil is_________ A -m². [Given π = 3.14]

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Vishal Baghel

Contributor-Level 10

^{$l = 2 π r$}

^{$\Rightarrow 314 c m = 2 * 3.14 r$}

^{$\Rightarrow r = \frac{1}{2} m = 0.5 m$}

^{$μ = i A$}

^{$= 14 * π r^{2}$}

^{$= 1 4^{2} * \frac{22}{7} * \frac{1}{4} = 11$}

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