The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. The solenoid has 2000 turns/m and its radius is 6.0 cm. (a) Consider a circle of with the axis of the solenoid. Write the change in the magnetic flux through this circle in 2.0 seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.

Question

Aditi Chauhan · Answer

Sol. di/dt = 0.01 A/s For 2s di/dt = 0.02 A/s n = 2000 turn/m, R = 6.0 cm = 0.06 m r = 1 cm = 0.01 m a) ϕ = BA ⇒ dϕ/dt = μ base 0 nA di/dt = 4π × 10^–7 × 2 × 10^3 × π × 1 × 10^–4 × 2 × 10^–2 [A = π × 1 × 10^–4] = 16π^2 × 10^–10 ω = 157.91 ×10^–10 ω = 1.6 × 10–8 ω or, dϕ/dt for 1 s=0.785 ω. (b) ∫▒〖E.〗dl = dϕ/dt ⇒ Eϕdl = dϕ/dt ⇒ E = 0.785x10^-8/2π x 10^-2 = 1.2 x 10^-7 V/m (c) dϕ/dt = μ base 0 n di/dt A = 4π x 10^-7 x 2000 x 0.01 x π x (0.06)^2 Eϕdl = dϕ/dt ⇒ E dϕ/dt/2πr = 4π x 10-7 x 2000 x 0.01 x π x (0.06)^2 = 5.64 x 10^-7 V/m

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

1 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Electromagnetic Induction

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship