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13 Electromagnetism I
13.54 For equilibrium, magnetic force = gravitational force.
F =
μ 0 li 1 i 2
2π d
= mg
∴ d =
μ 0 li 1 i 2
2π mg
=
4π × 10 −7 × 2 × 10 × 15
2π × 4 × 10 −3 × 9.8
= 1.53 × 10
−3 m
= 1.53 mm
13.55 (a) Net force per metre on the outer wire
F
l
=
F 1
l
+
F 2
l
=
μ 0 i 1 i 3
2 π d
+
μ 0 i 2 i 3
2 π × 2d
=
μ 0 i 3 (2i 1 + i 2 )
4π d
=
4π × 10 −7 × 20(2 × 20 + 20)
4π × 0.1
= 1.2 × 10
−3 N
(b) Zero
13.56 F = ilB + (i)(π R)B + ilB = i(2l + π R)B.
13.57 (i) The force on the horizontal segments is zero as they are perpendicular
to the straight wire. For the vertical segments the force is repulsive for
antiparallel currents and attractive for parallel currents, the magnitude
being
F =
μ 0 li 1 i 2
2π d
(1)
where d is the distance of separation.
Force on the nearer vertical segment
F 1 = −
(4π × 10 −7 )(0.1)(20)(5)
(2π)(0.02)
= −1 × 10
−4 N
Force on the farther vertical segment
F 2 = +
(4π × 10 −7 )(0.1)(20)(5)
(2π)(0.07)
= + 2.86 × 10
−5 N
(ii) The net force on the coil, F net = F 1 + F 2 = −7.14 × 10 −5 N
13.58 (a) Wire PQ will produce a field of induction B 1 at the segment AB of the
coil. The magnitude of B 1 will be
B 1 =
μ 0 I 1
2π d
(1)
The right-hand rule shows that the direction of B 1 at the segment AB
is down, wire AB, which carries a current I 2 finds itself immersed in
13 Electromagnetism I
13.54 For equilibrium, magnetic force = gravitational force.
F =
μ 0 li 1 i 2
2π d
= mg
∴ d =
μ 0 li 1 i 2
2π mg
=
4π × 10 −7 × 2 × 10 × 15
2π × 4 × 10 −3 × 9.8
= 1.53 × 10
−3 m
= 1.53 mm
13.55 (a) Net force per metre on the outer wire
F
l
=
F 1
l
+
F 2
l
=
μ 0 i 1 i 3
2 π d
+
μ 0 i 2 i 3
2 π × 2d
=
μ 0 i 3 (2i 1 + i 2 )
4π d
=
4π × 10 −7 × 20(2 × 20 + 20)
4π × 0.1
= 1.2 × 10
−3 N
(b) Zero
13.56 F = ilB + (i)(π R)B + ilB = i(2l + π R)B.
13.57 (i) The force on the horizontal segments is zero as they are perpendicular
to the straight wire. For the vertical segments the force is repulsive for
antiparallel currents and attractive for parallel currents, the magnitude
being
F =
μ 0 li 1 i 2
2π d
(1)
where d is the distance of separation.
Force on the nearer vertical segment
F 1 = −
(4π × 10 −7 )(0.1)(20)(5)
(2π)(0.02)
= −1 × 10
−4 N
Force on the farther vertical segment
F 2 = +
(4π × 10 −7 )(0.1)(20)(5)
(2π)(0.07)
= + 2.86 × 10
−5 N
(ii) The net force on the coil, F net = F 1 + F 2 = −7.14 × 10 −5 N
13.58 (a) Wire PQ will produce a field of induction B 1 at the segment AB of the
coil. The magnitude of B 1 will be
B 1 =
μ 0 I 1
2π d
(1)
The right-hand rule shows that the direction of B 1 at the segment AB
is down, wire AB, which carries a current I 2 finds itself immersed in