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 circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.1 T normal to the plane of the coil. If the current in the coil is 5.0 A What is the average force on each electron in the coil due to the magnetic field ( the coil is made of copper wire of cross - sectional area 10 to power minus five m square and the free electron density in copper is given to be about 10 power 29 m to power minus three.

A circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.1 T normal to the plane of the coil. If the current in the coil is 5.0 A What is the average force on each electron in the coil due to the magnetic field ( the coil is made of copper wire of cross - sectional area 10 to power  minus five m square and the free electron density in copper is given to be about 10 power 29 m to power minus three.

Grade:10

1 Answers

Arun
25763 Points
2 years ago

Number of turns on the circular coil, n = 20

Radius of the coil, r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

(a) The total torque on the coil is zero because the field is uniform.

(b) The total force on the coil is zero because the field is uniform.

(c) Cross-sectional area of copper coil, A = 10−5 m2

Number of free electrons per cubic meter in copper, N = 1029 /m3

Charge on the electron, e = 1.6 × 10−19 C

Magnetic force, F = Bevd

Where,

vd = Drift velocity of electrons

 

 = I / NeA

 

F = 5 * 10^-25 N

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