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Magnetic !elds (/learn/magnetic-!elds-questionsand-answers.html)
The circuit below shown below is made of a
battery, a pair of resistors, and a metal bar
which...
Question:
direction.
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The circuit below shown below is made of a battery, a pair of resistors, and a metal
bar which rests on the wires and has a mass of 560 grams and 2.5 meters long and
has an internal resistance of 11.5 Omega. The bar is a completely uniform magnetic
!eld 3.5 Tesla in magnitude and oriented to the right. The bar is free to move in any
(a) When the switch is closed, what is the current "owing through the bar?
(b) What is the magnitude of the acceleration the bar experiences after the switch is
closed?
(c) What is the direction of the acceleration the bar experiences after the switch is
closed?
(d) Assuming the force of gravity is directed into the paper, what is the maximum
height the bar goes to?
"
Magnetic Force in a Current Carrying Conductor:
When an external magnetic !eld acts on a current carrying conductor, then it
experiences a force.
This force is represented by the vector
→)
F→ = L. (→i × B
The direction of the force is given by the cross product of the direction of current and
the external magnetic !eld.
L is the length of the conductor.
#
Answer and Explanation:
Given;
Mass of the wire: m
= 0.56 Kg
Length of the bar: L
= 2.5 m
= 3.5
Magnetic !eld: B
= 3.5 T
Internal resistance of the bar: R
Resistors: R1
Battery emf: V
= 11.5 Ω
= R2 = 20 Ω
= 240 V
In the circuit, the resistors R1 and R2 are in parallel and the combination is in series
with R.
Total e#ective resistance of the circuit is:
Req =
R1 × R2
20 × 20
+ R =
+ 11.5 = 21.5 Ω
R1 + R2
40
PART (a)
Current through the circuit is:
I =
V
240
=
= 11.2 A
Req
21.5
PART (b)
Magnitude of force on the bar due to magnetic !eld is (current and the magnetic !eld
are in perpendicular directions):
FB = B ⋅ I ⋅ L
= 3.5 × 11.16 × 2.5 N
= 97.7 N
(Current is downwards and the magnetic !eld is to the right. Therefore, from the righthand rule, the force on the bar will be out of the page).
Force due to gravity:
Fg = m × g = 0.56 × 9.8 = 5.5 N ( into the page)
Net force on the bar is:
F = FB − Fg
= 97.7 N − 5.5 N
= 92.2 N
(out of the plane of the page)
Acceleration experienced by the bar is:
F
m
92.2
=
m/s2
0.56
= 164.6 ms−2
a =
PART (c).
The direction of acceleration will be in the direction of net force i.e. out of the plane
of page.
PART (d).
The potential on the 11.5Ω is given by ϵ = i × R = 11.2 × 11.5 V = 128.8 V .
Thus from the expression of motional EMF we have the initial velocity v of the bar
ϵ = BvL
128.8 = 3.5 × v × 2.5
v = 14.7 m/s
Thus the maximum height the bar goes to is given by the equation of motion as
v2f = v2 + 2as
0 = 14.72 + 2 × (−164.6) × h
h = 0.656 m