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Physics Moving Charges and Magnetism

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3 months 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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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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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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3 months 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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3 months ago

Physics Moving Charges and Magnetism Moving Charges and Magnetism

A closely wounded circular coil of radius 5cm produces a magnetic field of 37.68 * 10^-4 T at its centre. The current through the coil is_ A. [Given, number of turns in the coil is 100 and π = 3.14]

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Vishal Baghel

Contributor-Level 10

$B_{c e n t r e} = N \frac{μ_{0} i}{2 r}$

$\Rightarrow \frac{100 * 4 π * 1 0^{- 7} * i}{2 * 5 * 1 0^{- 2}} = 37.68 * 1 0^{- 4}$

$∴ i = \frac{37.68}{4 * 3.14} = 3 A$

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3 months ago

Physics Moving Charges and Magnetism Class 12th

A proton, a deuteron and an α - particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B} = \frac{\sqrt[]{2 m k . E}}{q B}$

$r \propto \frac{\sqrt[]{m}}{q}$

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_{p}}{r_{d}} = \frac{\sqrt[]{m_{p}} * q_{d}}{q_{P} * \sqrt[]{m_{d}}}$

$\frac{r_{d}}{r_{α}} = \sqrt[]{2}$

$r_{p} : r_{d} : r_{α} = 1 : \sqrt[]{2} : 1$

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Physics Moving Charges and Magnetism Moving Charges and Magnetism

The magnitude field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is :

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Vishal Baghel

Contributor-Level 10

$B_{C} = \frac{μ_{0} l}{2 r} = B$ $_{} \frac{_{}}{}$ -(1)

$\begin{array}{l} B_{P} = \frac{μ_{0} l r^{2}}{2 {(r^{2} + \frac{r^{2}}{4})}^{3 / 2}} \\ B_{P} = \frac{μ_{0} l r^{2}}{2 * r^{3} {(\frac{5}{4})}^{3 / 2}} \end{array}$

$B_{P} = \frac{μ_{0} l}{2 r {(\frac{5}{4})}^{3 / 2}}$

$\Rightarrow B_{P} = \frac{B 4^{3 / 2}}{5^{3 / 2}}$

$B_{P} = B \frac{2^{3}}{{(\sqrt[]{5})}^{3}}$

$B_{P} = {(\frac{2}{\sqrt[]{5}})}^{3} B$

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Physics Moving Charges and Magnetism Class 12th

Given below are two statements : One is labeled as Assertion (A) and the other is labeled as Reason (R). &nbs

...more

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Vishal Baghel

Contributor-Level 10

Work done in magnetic field zero.

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Physics Moving Charges and Magnetism Class 12th

Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $μ T$ . the equal current flowing in the two conductors is

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Payal Gupta

Contributor-Level 10

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_{P} = \frac{μ_{0} I}{2 π r} * 2$

$\Rightarrow 300 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π * 4 * 1 0^{- 2}} * 2$

$\Rightarrow$ I = 30 A

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3 months ago

Physics Moving Charges and Magnetism Class 12th

In a Young's double slit experiment, an angular width of the fringe is 0.53° on a screen place at 2m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is $\frac{1}{α} .$ The value of a is__________.

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alok kumar singh

Contributor-Level 10

we know, = $\frac{λ}{d}$ $μ = \frac{C}{V} = \frac{C}{υ λ}$

$λ_{1} = \frac{C}{μ_{1} υ_{1}}$ [with change in medium frequency refnain constaint]

$λ_{2} = \frac{C}{μ_{2} υ_{2}}$

$\frac{λ_{2}}{λ_{1}} = \frac{μ_{1}}{μ_{2}} = \frac{5}{7}$ $λ_{2} = \frac{5}{7} λ_{1}$

$θ_{1} = \frac{λ_{1}}{d_{1}}$

2 = $\frac{λ_{2}}{d_{2}}$

$\frac{θ_{2}}{θ_{1}} = \frac{λ_{2}}{λ_{1}} [? d_{1} = d_{2}]$

2 $= \frac{5}{7} * θ_{1}$

$θ_{2} = \frac{5}{7} * 0.35$

$θ_{2} = \frac{1}{4} = \frac{1}{α}$

= 4

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3 months ago

Physics Moving Charges and Magnetism Class 12th

At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter. 6.241cm? Both the diameters have been measured at room temperature (27°C).

(Given : coefficient of linear thermal expansion of gold a_L = 1.4 * 10^-⁵ K^-¹)

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alok kumar singh

Contributor-Level 10

According to question, we can write

$d = d_{0} (1 + α Δ T) \Rightarrow 6.241 = 6.230 (1 + 1.4 * 1 0^{- 5} * Δ T)$

$\Rightarrow T = 126.18 + 27 = 153.18 ° C$

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