drive the expression for torque exerted on bar magnet placed in mag. field
Answers
Answer:
Given: Torque = τ = 2 x 10-5 Nm, Magnetic induction = B = 2 x 10-5 Wb/m2, angle with field = θ = 90°.
Explanation:
i hope this helps you a lot
Answer:
ANSWER
Let NS be a bar magnet of pole strength m and effective length 2l. Consider a point P on the neutral axis at a distance d from the centre of the magnet.
Let PN=PS=x.
Now, the intensity of field at P due to N-pole is,
B
1
=
4π
μ
0
⋅
x
2
m
(along
NP
)
Similarly, the intensity of field at P due to S pole is given by
B
2
=
4π
μ
0
⋅
x
2
m
(along
PS
)
Since, B
1
=B
2
=
4π
μ
0
⋅
x
2
m
Hence, B
1
=B
2
Hence, by the law of parallelogram of force the resultant will be given by
B=
B
1
2
+B
2
2
+2B
1
B
2
cos2θ
(Where 2Q is the angle between B
1
and B
2
)
But, B
1
=B
2
, hence
B=
B
1
2
+B
1
2
+2B
1
2
cos2θ
=
2B
1
2
+2B
1
2
cos2θ
=
2B
1
2
(1+cos2θ)
=
2B
1
2
[1+2cos
2
θ−1]
=
2B
1
2
⋅2cos
2
θ
=2B
1
cosθ
or B=2⋅
4π
μ
0
⋅
x
2
m
cosθ
or B=
4π
μ
0
⋅
x
2
2m
⋅
x
1
,(∵cosθ=
x
1
)
or B=
4π
μ
0
⋅
x
3
2ml
But, 2ml=M (magnetic moment)
∴B=
4π
μ
0
⋅
x
3
2M
But, by the Pythagoras theorem in right angle ΔPQS
x=
d
2
+1
2
or x
3
=(d
2
+1
2
)
3/2
Hence, B=
4π
μ
0
⋅
(d
2
+1
2
)
3/2
M
If the magnet is small i.e. d≫1
∴B=
4π
μ
0
⋅
d
3
M
This is the required expression.
The direction of magnetic field is from north pole to south pole and parallel to magnetic axis of the magnet.
solution
