Question

A wire of length L is bent into a square loop and is placed in a magnetic field to produce a motor. For a fixed value of current and magnetic field, what will produce a greater torque, forming the wire into a single large loop or two, smaller loops?
It depends on the exact values of I and B.
the larger loop.
two smaller loops.
It is the same.

Answer 1

@shivam_bhalla do u know how can i solve it please :)

Answer 2

For a circular ring,
\[\large B = \frac{\mu_0 n I }{2r}\]
n-->number of loops
I--> current in loop
r-->radius of loop

Now think carefully and try the question
Hopefully you know
L = 2 (pi)(r)

Now Take case 1 --> One loop
Take case 2 -->Two loops
and find radius for each case

Answer 3

Where L -->length of wire

For n loops
L = n (2)pi(r)

Now try :d

Answer 4

thanx :)

Answer 5

Try and tell me. I again repeat be careful while calculating radius of single loop and double loop

Answer 6

the answer will be ( It is the same.)

Answer 7

@shivam_bhalla my answer is correct or not ?

Answer 8

let me check

Answer 9

Well If forgot about the torque. For torque,

t = B I A n cosq

where B = magnetic flux density of the field, I = current flowing in coil, A = area of coil, n = number of turns of wire in coil and q = initial angle made by plane of coil and the B field direction.
Note that when q = 90o (coil perpendicular to field direction), no torque exits since then F and d are both in the same plane.

Answer 10

*I

Answer 11

ok my answer now is wrong

Answer 12

Well, you can always correct your answer @blacktigger

Answer 13

i asked my friend he told me it will be the larger loop. @shivam_bhalla

Answer 14

I am sorry @ Blacktigger. I have made very foolish mistakes I have complicated the problem. The mistake I have made is that I assumed it as circle


It should be
torque, t = B I A n cosq
where B = magnetic flux density of the field, I = current flowing in coil, A = area of coil, n = number of turns of wire in coil and q = initial angle made by plane of coil and the B field direction. Note that when q = 90o (coil perpendicular to field direction), no torque exits since then F and d are both in the same plane.

Area of coil = a^2

a-->side of a square coil
Now write

for A-->area of coil
A = a^2

Find a (side of square ) by using

For loop 1
L = 4a_1

For loop 2
L = 8a^2

Now try you should get the answer and sorry for the mistake I made

Answer 15

@blacktigger

Answer 16

ok np i will try it again now :)