APNI KAKSHA 1
, Moving Charges and Magnetism
Introduction
• Oersted experimentally demonstrated that a current in a straight wire caused deflection in a
nearby magnetic compass needle.
• Moving charges or currents produce a magnetic field in the surrounding space.
• we adopt the following convention: A current or a field (electric or magnetic) emerging out
of the plane of the paper is depicted by a dot (⊙). A current or a field going into the plane of
the paper is depicted by a cross (⊗).
Magnetic field
• Magnetic field is the space around a moving charge or a magnetic material in which its
magnetic influence can be experienced.
• It is a vector quantity.
• It's SI unit is Tesla (= weber /m2 ). [1 gauss = 10−4 Tesla].
• It's dimensional formula is given by [MT −2 A−1 ].
• Principle of superposition: the magnetic field of several sources is the vector addition of
magnetic field of each individual source.
Magnetic force
• Consider a charge q moving with velocity V in a magnetic field B, then magnetic force on
⃗ B = q(V
charge q is given by F ⃗ ×B
⃗)
• It depends on q, V and B. Force on a negative charge is opposite to that on a positive charge.
• The magnetic force is zero if charge is not moving (as then |V| = 0 ) or velocity and magnetic
field are parallel or antiparallel.
• Magnetic force is perpendicular to both velocity and magnetic field. Its direction is given by
the screw rule or right hand rule for vector (or cross) product.
Magnetic force on a current – carrying conductor
Consider a rod of a uniform cross-sectional area A and length ℓ. Let the number density of these
mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nℓA.
let drift velocity be Vd . then total magnetic force on the rod in presence of external magnetic field
is F = (nℓA)eVd B
⇒ F = (neAVd )ℓB = IℓB (∵ I = ne AVd )
in vector form; F ⃗ ×B
⃗ = I(ℓ ⃗)
where ⃗ℓ is a vector of magnitude ℓ, the length of the rod, and with a direction identical to the
current ℓ.
APNI KAKSHA 2
, Q. A straight wire of mass 𝟐𝟎𝟎 𝐠 and length 𝟏. 𝟓 𝐦 carries a current of 𝟐 𝐀. It is
suspended in mid-air by a uniform horizontal magnetic field 𝐁. What is the
magnitude of the magnetic field? [NCERT Exercise]
Sol. We can say that there is an upward force F, of magnitude I𝑙B,. For mid-air suspension,
this must be balanced by the force due to gravity:
mg = I𝑙B
mg 0.2 × 9.8
B= = = 0.65T
I𝑙 2 × 1.5
Motion in a magnetic field
(i) Let a charge particle of mass m enters in a magnetic field of magnitude B. First consider the
case of v perpendicular to B. The perpendicular force, qv × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field. The particle will describe a
circle if v and B are perpendicular to each other (Fig.). Let radius of circle is r.
mv 2
Therefore; q∨B=
r
mv
⇒ r=
qB
Time period of revolution,
2πr 2πm
T= =
v qB
Angular Frequency,
2π qB
ω= =
T m
(ii) Now consider velocity makes an angle θ with magnetic field. Here velocity, will have a
component along B and this component remains unaffected due to magnetic field. Motion in
the plane perpendicular to the B is circular because of component of velocity perpendicular
to B, thereby producing a helical motion.
For circular motion;
m(vsinθ)2
q(vsin θ)B =
r
mvsin θ
r=
qB
2πr 2πm
Time period = T = =
vsinθ qB
APNI KAKSHA 3
, Moving Charges and Magnetism
Introduction
• Oersted experimentally demonstrated that a current in a straight wire caused deflection in a
nearby magnetic compass needle.
• Moving charges or currents produce a magnetic field in the surrounding space.
• we adopt the following convention: A current or a field (electric or magnetic) emerging out
of the plane of the paper is depicted by a dot (⊙). A current or a field going into the plane of
the paper is depicted by a cross (⊗).
Magnetic field
• Magnetic field is the space around a moving charge or a magnetic material in which its
magnetic influence can be experienced.
• It is a vector quantity.
• It's SI unit is Tesla (= weber /m2 ). [1 gauss = 10−4 Tesla].
• It's dimensional formula is given by [MT −2 A−1 ].
• Principle of superposition: the magnetic field of several sources is the vector addition of
magnetic field of each individual source.
Magnetic force
• Consider a charge q moving with velocity V in a magnetic field B, then magnetic force on
⃗ B = q(V
charge q is given by F ⃗ ×B
⃗)
• It depends on q, V and B. Force on a negative charge is opposite to that on a positive charge.
• The magnetic force is zero if charge is not moving (as then |V| = 0 ) or velocity and magnetic
field are parallel or antiparallel.
• Magnetic force is perpendicular to both velocity and magnetic field. Its direction is given by
the screw rule or right hand rule for vector (or cross) product.
Magnetic force on a current – carrying conductor
Consider a rod of a uniform cross-sectional area A and length ℓ. Let the number density of these
mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nℓA.
let drift velocity be Vd . then total magnetic force on the rod in presence of external magnetic field
is F = (nℓA)eVd B
⇒ F = (neAVd )ℓB = IℓB (∵ I = ne AVd )
in vector form; F ⃗ ×B
⃗ = I(ℓ ⃗)
where ⃗ℓ is a vector of magnitude ℓ, the length of the rod, and with a direction identical to the
current ℓ.
APNI KAKSHA 2
, Q. A straight wire of mass 𝟐𝟎𝟎 𝐠 and length 𝟏. 𝟓 𝐦 carries a current of 𝟐 𝐀. It is
suspended in mid-air by a uniform horizontal magnetic field 𝐁. What is the
magnitude of the magnetic field? [NCERT Exercise]
Sol. We can say that there is an upward force F, of magnitude I𝑙B,. For mid-air suspension,
this must be balanced by the force due to gravity:
mg = I𝑙B
mg 0.2 × 9.8
B= = = 0.65T
I𝑙 2 × 1.5
Motion in a magnetic field
(i) Let a charge particle of mass m enters in a magnetic field of magnitude B. First consider the
case of v perpendicular to B. The perpendicular force, qv × B, acts as a centripetal force and
produces a circular motion perpendicular to the magnetic field. The particle will describe a
circle if v and B are perpendicular to each other (Fig.). Let radius of circle is r.
mv 2
Therefore; q∨B=
r
mv
⇒ r=
qB
Time period of revolution,
2πr 2πm
T= =
v qB
Angular Frequency,
2π qB
ω= =
T m
(ii) Now consider velocity makes an angle θ with magnetic field. Here velocity, will have a
component along B and this component remains unaffected due to magnetic field. Motion in
the plane perpendicular to the B is circular because of component of velocity perpendicular
to B, thereby producing a helical motion.
For circular motion;
m(vsinθ)2
q(vsin θ)B =
r
mvsin θ
r=
qB
2πr 2πm
Time period = T = =
vsinθ qB
APNI KAKSHA 3
