Physics Moving Charges and Magnetism

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New answer posted•a year 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

19 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

Force on Y = 1.2 × 105N towards ‘x’

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New answer posted•a year 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:

18 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year 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:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year ago

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:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

T₁ = 3 sec T₂ = 4sec

$T = 2 π \sqrt[]{\frac{I}{μ B}}$

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

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

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New answer posted•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)

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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