Chapter 4: Q39E (page 984)

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the centre of the solenoid and 3.50 cm from its axis is 8.00 x 10-6 V/m. Calculate di/dt

Short Answer

Expert verified

If a long thin solenoid having 400 turns per meter and radius of , the current in the solenoid is increasing at a unform rate of 9.21A/s

Step by step solution

Calculating the value of E

It is given that a long solenoid of 400 turns and distance of $m^{- 1}$ and a radius of 1.10cm . The induced electric field at a point near the centre of the solenoid is 3.50cm and induced electric field is $8.00 \times 10^{- 6} V / m$

Now using Faraday’s law we get:

$ε = \oint \vec{E} . \vec{d l} = - \frac{d ϕ B}{d t}$

Now, the magnetic flux through the circle of radius r > R :

$ϕ = AB = {πR}^{2} B$

Now integrating LHS we get:

$\oint \vec{E} . \vec{d l} = E (2 π r)$

Absolute value of E is:

$E (2 π r) = π R^{2} |\frac{d B}{d t}|$

We can also write it as:

$E = \frac{R^{2} μ_{0} n}{2 r} = |\frac{d B}{d t}|$

Calculating the values

Now to get $\frac{d l}{d t}$ we get:

$|\frac{dl}{dt}| = \frac{2 Er}{μ_{0} {nR}^{2}}$

Now, substituting the values we get:

$\frac{2 (8.00 \times 10^{- 6} V / m) (0.0350 m)}{(4 π \times 10^{- 7} T . m / A) (400) {(0.0110 m)}^{2}} = 9.21 A / s$

Therefore, the value of $|\frac{d l}{d t}|$ is = 9.21A/s

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A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the centre of the solenoid and 3.50 cm from its axis is 8.00 x 10-6 V/m. Calculate di/dt

Short Answer

Step by step solution

Calculating the value of E

Calculating the values

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