Question

A solenoid of 40 cm length has 600 turns and its circular cross section

Answer 1

24000 is its circular section that is senoid sectiom

Answer 2

Question:

A solenoid of 40 cm length has 600 turns and its circular cross section has a diameter of 4 cm. The rate of change of current in the solid will produce a self induced emf of 0.4 V is ?

Answer:

The rate of change of producing self induced emf of 0.4 V is 281.5 A/s.

Explanation:

Given,

The length of the solenoid = 40 cm = 0.4 m

Turns in the solenoid = 600 turns

Diameter (cross section) = 4 cm = 4 × 10⁻² m

Self induced emf = 0.4V

To Find:

The rate of change of current in the solid will produce a self induced emf of 0.4 V is:

Formulae:

$Inductance , L =$ μ₀ × $\frac{N^{2} A }{l}$  .......................................(1)

To find the value of A,

$A = \frac{\pi d^{2} }{4}$   .................................................(2)

Solution:

A = $\frac{\pi }{4}$ × ( 4 × 10⁻²)² m²

   = $\frac{\pi }{4}$ × ( 16 × 10⁻⁴) m²

 A = 4π × 10⁻² m²

Therefore,  

$V = L \frac{di}{dt}$     or  

$\frac{di}{dt} = \frac{V}{L} = \frac{V L}{N^{2} A }$× 1/ μ₀ ..................(3)

Substitute the values in equation (3), we get

$\frac{di}{dt} = \frac{0.4 * 0.4}{4 \pi * 10^{-7 }(600^{2})(4\pi *10^{-4} }$   ...................................(4)

By solving the above equation we get the answer,

$\frac{di}{dt} = 281.5 A/s$

Answer = The rate of change of producing self induced emf of 0.4 V is 281.5 A/s.

#SPJ2