Physics Moving Charges and Magnetism

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New answer posted•a year 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]

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

$∴ i = \frac{37.68}{4 \times 3.14} = 3 A$

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New answer posted•a year 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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

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

$\frac{r_{d}}{r_{α}} = \sqrt[]{2}$

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

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New answer posted•a year ago

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 :

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

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

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New answer posted•a year ago

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). (Magnetism)

Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle.

Reason (R) : Moving charged particle experiences magnetic force perpendicular to its direction of motion.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Work done in magnetic field zero.

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New answer posted•a year ago

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

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$B_{P} = \frac{μ_{0} I}{2 π r} \times 2$

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

$\Rightarrow$ I = 30 A

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New answer posted•a year 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__________.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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 $= \frac{5}{7} \times θ_{1}$

$θ_{2} = \frac{5}{7} \times 0.35$

$θ_{2} = \frac{1}{4} = \frac{1}{α}$

= 4

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New answer posted•a year 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)

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

According to question, we can write

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

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

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 × 10^-6N, then the value of x is approximately:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{F}{l} = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{d}$

$2 \times 1 0^{- 6} = \frac{4 π \times 1 0^{- 7}}{2 π} \times \frac{x \times x}{d}$

x² = 2x = $\sqrt[]{2}$ = 1.4x = 1.4

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New answer posted•a year ago

Physics Moving Charges and Magnetism Class 12th

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 × 10^-5. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$χ (S u s c e p t i b i l i t y) = 1.2 \times 1 0^{- 5}$

fractional change = $= \frac{Δ B}{B} = \frac{B i n s i d e - B o u t}{B o u t}$

$= \frac{μ_{m} H - μ_{0} H}{μ_{0} H} = \frac{(μ_{0} μ_{r} - μ_{0}) H}{μ_{0} H}$ $[? μ_{m} = b_{0} μ_{0}]$

$= μ_{r} - 1$

= $= (1 + χ) - 1 = χ = 1.2 \times 1 0^{- 5}$

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New question posted•a year ago

Physics Moving Charges and Magnetism Class 12th

<p><strong>A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:</strong></p>

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Physics Moving Charges and Magnetism

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