Question

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equation B(t) = 5.0t T, where t is time in seconds. If the induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, find the radius of the solenoid.

Answer 1

The radius of the solenoid is 0.94 meters.

Step-by-step explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation as :

`B(t)=0.5t\ T`

Where

t is the in seconds

The induced electric field outside the solenoid is,
`\epsilon=1.1\ V/m`

Distance, d = 2 m from the axis of the solenoid

To find,

The radius of the solenoid.

Solution,

`B(t)=0.5t\ T`

`(dB)/(dt)=5\ T`

The expression for the induced emf is given by :

`\epsilon=(d\phi)/(dt)`

`\phi` = magnetic flux

The electric field due to changing magnetic field is given by :

`\epsilon(2\pi x)=(d(BA))/(dt)`

`\epsilon(2\pi x)=\pi r^2(d(B))/(dt)`

`r^2=(2\epsilon x)/(dB/dt)`

`r^2=(2* 1.1* 2)/(5)`

r = 0.9380 m

or

r = 0.94 meters

So, the radius of the solenoid is 0.94 meters. Hence, this is the required solution.

Answer 2

Radius of the solenoid is 0.93 meters.

Step-by-step explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

`B(t)=5t\ T`, t is time in seconds

`(dB)/(dt)=5\ T`

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

`E(2\pi x)=(d\phi)/(dt)`

x is the distance from the axis of the solenoid

`E(2\pi x)=\pi r^2(dB)/(dt)`, r is the radius of the solenoid

`r^2=(2xE)/((dE/dt))`

`r^2=(2* 2* 1.1)/((5))`

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.