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

The magnetic flux near the axis inside an air core solenoid carrying a current I is π×10^-6/2 Wb. The length the solenoid is 60cm. What will be its magnetic moment ? (Lengt solenoid h of solenoid is large as compared with its cross section)

The magnetic flux near the axis inside an air core solenoid carrying a current I is π×10^-6/2 Wb. The length the solenoid is 60cm. What will be its magnetic moment ? (Lengt solenoid h of solenoid is large as compared with its cross section)

Grade:12

1 Answers

Manisha Kumari
23 Points
2 years ago
   B=uoNI/L   flux =B.A   A=flux/B  M=NIA = NI×flux×L/uoNI= flux × L/uo       =π×10^-6×60×10^-2/2×4π×10^-7=0.75 Am^2

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