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Allisa [31]
3 years ago
8

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

ion 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.
Physics
2 answers:
Contact [7]3 years ago
5 0

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.

andrezito [222]3 years ago
4 0

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.

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