Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A toroird having a square cross section , 5.20cm on edge , and an inner radius of 16.2cm has 535 turns and carries a current of 813mA calulate the magnetic field inside the toroid at (a) the inner and (b) the outer radius of the toroid

A toroird having a square cross section , 5.20cm on edge , and an inner radius of 16.2cm has 535 turns and carries a current of 813mA calulate the magnetic field inside the toroid at (a) the inner and (b) the outer radius of the toroid

Grade:12th pass

1 Answers

Vikas TU
14149 Points
3 years ago
Using the equation, B = (μoIN/2πr)
The inner radius is r = 16.2 cm, so the field there is
B = (4π 10-7).(0.813).(535)/2π(0.162)
= 5.37 × 10-4 T
 
The outer radius is r = 16.2 + 5.2 = 21.4 cm.
The field there is B = (4π 10-7).(0.813).(535)/2π(0.214)
= 4.06 × 10-4 T
 

Think You Can Provide A Better Answer ?

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

ASK QUESTION

Get your questions answered by the expert for free