Question

A solenoid has 2 x 10 ^4 turns/ meter and has diameter 10 cm. An electron beam having ke

Answer 1

The total charge flowing through the coil during this time is q = 32 μC

Explanation:

Correct statement:

A long solenoid of diameter 0.1 m has 2×104 turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A from 4 a in 0.05 s . If the resistance of the coil is 10π2Ω, the total charge flowing through the coil during this time is

Solution:

q = [(ΔϕΔt)1 / R]Δt

q = [μ0nNπr^2Δi/Δt]1/RΔt

q = [4×10^−7×2×10^4×100×π×(10^−2)^2×(40.05)] 1 /10π^2×0.05

q = [ 2512  x 10-7+4 -4 x 4 / 0.05] 1/10 x 3.14 x 3.14 x 0.05

q = [ 2.5 x 10^-4 x 4/ 0.05 ] x 1/ 98.59 x 0.05

q = 32 μC

Thus the total charge flowing through the coil during this time is q = 32 μC

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Answer 2

The current in the solenoid is 0.421 A.

Explanation:

Given that,

Diameter = 10 cm

Suppose an electron beam having K.E 100 keV passes without touching walls of solenoid then find current in the solenoid. Electron beam make angle 30° with axis of solenoid.

Let v is the speed of the electron.

We need to calculate the velocity

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$v^2=\dfrac{2\times K.E}{m}$

Put the value into the formula

$v^2=\dfrac{2\times100\times10^{3}\times1.6\times10^{-19}}{9.1\times10^{-31}}$

$v=\sqrt{\dfrac{2\times100\times10^{3}\times1.6\times10^{-19}}{9.1\times10^{-31}}}$

$v=187522892.375\ m/s$

$v=1.87\times10^{8}\ m/s$

let B is the magnetic field inside the solenoid

We need to calculate the magnetic field

Using formula of radius of path of the electron

$r=\dfrac{mv\sin\theta}{Bq}$

$B=\dfrac{mv\sin\theta}{qr}$

Put the value into the formula

$B=\dfrac{9.1\times10^{-31}\times1.875\times10^{8}\sin30}{1.6\times10^{-19}\times5\times10^{-2}}$

$B=0.0106\ T$

We need to calculate the current in the solenoid

Using formula of magnetic field

$B=\mu_{0}\times n\times I$

$I=\dfrac{B}{\mu_{0} n}$

Put the value into the formula

$I=\dfrac{0.0106}{4\pi\times10^{-7}\times2\times10^{4}}$

$I=0.421\ A$

Hence, The current in the solenoid is 0.421 A.

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Topic : magnetic field

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