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

Answer:

The radius of the solenoid is 0.94 meters.

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,

$B(t)=0.5t\ T$

$\dfrac{dB}{dt}=5\ T$

The expression for the induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

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

$\epsilon(2\pi x)=\dfrac{d(BA)}{dt}$

$\epsilon(2\pi x)=\pi r^2\dfrac{d(B)}{dt}$

$r^2=\dfrac{2\epsilon x}{dB/dt}$

$r^2=\dfrac{2\times 1.1\times 2}{5}$

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

Answer:

Radius of the solenoid is 0.93 meters.

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

$\dfrac{dB}{dt}=5\ T$

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 :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$, r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

$r^2=\dfrac{2\times 2\times 1.1}{(5)}$

r = 0.93 meters

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