Question

What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic field of 3.0 T/s causes a 6.0 A current to flow?The resistance of the circuit that contains the solenoid is 17 ȍ. The only emf source forthe circuit is the induced emf.

Answer 1

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, $\dfrac{dB}{dt}=3\ T/s$

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

$E=N\dfrac{d\phi}{dt}$

$E=N\dfrac{d(BA)}{dt}$

Since, E = IR

$IR=NA\dfrac{dB}{dt}$

$A=\dfrac{IR}{N.\dfrac{dB}{dt}}$

$A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}$

$A=0.188\ m^2$

or

$A=0.19\ m^2$

Area of circular cross section is, $A=\pi r^2$

$r=\sqrt{\dfrac{A}{\pi}}$

$r=\sqrt{\dfrac{0.19}{\pi}}$

r = 0.24 m

So, the  radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.