Respuesta :
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,
[tex]B(t)=5t\ T[/tex], t is time in seconds
[tex]\dfrac{dB}{dt}=5\ T[/tex]
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 :
[tex]E(2\pi x)=\dfrac{d\phi}{dt}[/tex]
x is the distance from the axis of the solenoid
[tex]E(2\pi x)=\pi r^2\dfrac{dB}{dt}[/tex], r is the radius of the solenoid
[tex]r^2=\dfrac{2xE}{(dE/dt)}[/tex]
[tex]r^2=\dfrac{2\times 2\times 1.1}{(5)}[/tex]
r = 0.93 meters
So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.
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 :
[tex]B(t)=0.5t\ T[/tex]
Where
t is the in seconds
The induced electric field outside the solenoid is, [tex]\epsilon=1.1\ V/m[/tex]
Distance, d = 2 m from the axis of the solenoid
To find,
The radius of the solenoid.
Solution,
[tex]B(t)=0.5t\ T[/tex]
[tex]\dfrac{dB}{dt}=5\ T[/tex]
The expression for the induced emf is given by :
[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
The electric field due to changing magnetic field is given by :
[tex]\epsilon(2\pi x)=\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon(2\pi x)=\pi r^2\dfrac{d(B)}{dt}[/tex]
[tex]r^2=\dfrac{2\epsilon x}{dB/dt}[/tex]
[tex]r^2=\dfrac{2\times 1.1\times 2}{5}[/tex]
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.
