I1=4a,a1=2cm,distance between axise point to the center is 1m
I2=3A ,A2=2CM SAME DISTANCE
FIND THE NET MAGNETIC FIELD AT P
Answers
First we find B1 and B2 and find their resultant using vector notation.
Bnet = √(B1²+B2²)
B1 and B2 are actually magnetic field due to cicular coils
so for B1,
B1 = (4π*10^-7) I1/2a1
B1 = 4π*10^-7 x 4/2(2*10^-2)
B1 = 4π * 10^-5 T
similarly,
B2 = (4π*10^-7) I2/2a2
B2 = (4π*10^-7) x 3/2(2*10^-2)
B2 = 3π *10^-5 T
thus, Bnet = √{(3π*10^-5)²+(4π*10^-5)²}
thus on solving,
Bnet = 15.7*10^-5 T
Answer:
τ
=
M
×
B
=MBsinθ, here M and B are in the same direction so θ=0
∴
τ
=0
(ii)
We know F
B
=i
l
×
B
On line AB and CD magnetic forces are equal and opposite. So they cancel out each other.
Magnetic force on line AD is
F=i
l
×
B
=ilB
∵B=
2πr
μ
0
I
F=
2πr
μiIL
(attractive)
Magnetic force on line CB.
F=∣
F
∣=ilB
′
∵B
′
=
2πr
′
μ
0
I
F
′
=
2πr
μiIl
This force is repulsive
So net force is
F
n
=F−F
′
=
2π
mu
0
iIl
(
r
1
−
r
′
1
)
Now givei=4A, I=1A, r=1cm, r
′
=(1+2)=3 cm L=5 cm
Putting all the values in the above equation
F
n
=2×10
−7
×4×1×0.05[
0.01
1
−
0.03
1
]=26.66×10
−7
N