The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
Imagine both the charges located in a coordinate frame of reference. At A, amount of charge, qA = 2.5 x 10_7C At B, amount of charge, qB = —2.5 x 10-7 C Total charge of the system, q = qA + qB = 2.5 x 10_7 C - 2.5 x 10-7C = 0 Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 mRead more
Imagine both the charges located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 x 10_7C
At B, amount of charge, qB = —2.5 x 10-7 C
Total charge of the system, q = qA + qB = 2.5 x 10_7 C – 2.5 x 10-7C = 0
Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 m
Then Electric dipole moment of the system is given by,
p = qAxd = qBxd = 2.5 x 10_7 x 0.3 = 7.5 x 10~8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 x 10~8 C m along positive z-axis.
(a). Let us imagine two points A and B such that AO=OB and O is centre of line AB. The distance between the two charges is AB=20cm Therefore AO=OB=10cm Then net electric field at point O=E (say) Electric field at point O as a result of +3µC charge will be =E₁ (say) Where (1/4πε0) =9x 10⁹ and ε0 Read more
(a).
Let us imagine two points A and B such that AO=OB and O is centre of line AB.
The distance between the two charges is AB=20cm
Therefore AO=OB=10cm
Then net electric field at point O=E (say)
Electric field at point O as a result of +3µC charge will be =E₁ (say)
Where (1/4πε0) =9x 10⁹ and ε0 =Permittivity of free space
and OA =10cm =10x 10⁻²Nm 2C⁻²
Then E₁= (1/4πε0) x 3×10⁻⁶/(OA)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Electric field at point O as a result of -3µC charge will be =E2 (say)
Then magnitude of E₂(absolute value)= (1/4πε0) x ( -3×10⁻⁶)/(OB)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Therefore E= E₁+E2= 2 x (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.
Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹
=5.4 x 10⁶ NC⁻¹ along OB
Therefore, the electric field at mid-point O is 5.4 x 106 NC⁻¹ along OB.
(b).
A test charge of amount 1.5 x 10-9 C is placed at mid-point O.
q= 1.5 x 10⁻9C
Force experienced by the test charge = F (say)
Then F = qE
= 1.5 x 10⁻9 x 5.4 x 106 = 8.1 x 10-3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 x 10-3 N along OA.
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other. (b).If two field lines cross eachRead more
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
(b).If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
Predict the direction of induced current in the situations described by the following Figures. 6.18(a) to (f).
The direction of the induced current in a closed loop is given by Lenz's law, it states that: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The given pairs of figures show the direction of the induced current when tRead more
The direction of the induced current in a closed loop is given by Lenz’s law, it states that:
The polarity of induced emf is such that it tends to produce a current
which opposes the change in magnetic flux that produced it.
The given pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.
Using Lenz’s rule, the direction of the induced current in the given situations can be predicted as follows:
Ans (a).The direction of the induced current is along qrpq.
Ans (b).The direction of the induced current is along prqp.
Ans (c).The direction of the induced current is along yzxy.
Ans (d).The direction of the induced current is along zyxz.
Ans (e).The direction of the induced current is along xryx.
Ans (f).No current is induced since the field lines are lying in the plane of the closed loop.
See lessThe magnetic moment vectors µs and µl associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: µs = –(e/m) S, µl = –(e/2m)l Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Out of the two relations given, only one is in accordance with classical physics. The magnetic momemtum vector which as a result of orbital Angular Moment is given by , μl=−e/2m l It follows from the definitions ofμl and l. μl=iA=(−e/T)πr2 ...(i) Angular momentum, l=mvr=m(2πr/T)r ...(iiRead more
Out of the two relations given, only one is in accordance with classical physics.
The magnetic momemtum vector which as a result of orbital Angular Moment is given by ,
μl=−e/2m l
It follows from the definitions ofμl and l.
μl=iA=(−e/T)πr2 …(i)
Angular momentum, l=mvr=m(2πr/T)r …(ii)
where r is the radius of the circular orbit, which the electron of mass m and charge (−e) completes in time T.
Divide (i) by (ii), μl/l=[(−e/T)πr2 ]/(m(2πr/T)r )=−e/2m
∴μl=(−e/2m)l
Clearly μl and l will be antiparallel (both being normal to the plane of the orbit)
In contrast, μs/S=e/m. It is obtained on the basis of quantum mechanics.
See lessA system has two charges q₁ = 2.5 × 10⁻⁷C and q₂ = –2.5 × 10⁻⁷ C located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Imagine both the charges located in a coordinate frame of reference. At A, amount of charge, qA = 2.5 x 10_7C At B, amount of charge, qB = —2.5 x 10-7 C Total charge of the system, q = qA + qB = 2.5 x 10_7 C - 2.5 x 10-7C = 0 Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 mRead more
Imagine both the charges located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 x 10_7C
At B, amount of charge, qB = —2.5 x 10-7 C
Total charge of the system, q = qA + qB = 2.5 x 10_7 C – 2.5 x 10-7C = 0
Distance between two charges at points A and B, d= 15 + 15 = 30 cm = 0.3 m
Then Electric dipole moment of the system is given by,
p = qAx d = qBx d = 2.5 x 10_7 x 0.3 = 7.5 x 10~8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 x 10~8 C m along positive z-axis.
See lessTwo point charges q₁ = 3 μC and q₂ = –3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?
(a). Let us imagine two points A and B such that AO=OB and O is centre of line AB. The distance between the two charges is AB=20cm Therefore AO=OB=10cm Then net electric field at point O=E (say) Electric field at point O as a result of +3µC charge will be =E₁ (say) Where (1/4πε0) =9x 10⁹ and ε0 Read more
(a).
Let us imagine two points A and B such that AO=OB and O is centre of line AB.
The distance between the two charges is AB=20cm
Therefore AO=OB=10cm
Then net electric field at point O=E (say)
Electric field at point O as a result of +3µC charge will be =E₁ (say)
Where (1/4πε0) =9x 10⁹ and ε0 =Permittivity of free space
and OA =10cm =10x 10⁻²Nm 2C⁻²
Then E₁= (1/4πε0) x 3×10⁻⁶/(OA)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Electric field at point O as a result of -3µC charge will be =E2 (say)
Then magnitude of E₂(absolute value)= (1/4πε0) x ( -3×10⁻⁶)/(OB)²
= (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Therefore E= E₁+E2= 2 x (1/4πε0) x 3×10⁻⁶/(10x 10⁻²)² along OB
Since E₁& E2 have electric field in the same direction it will add-up and since the magnitude of both are equal we can just double it.
Therefore E=2x 9x 10⁹ x 3×10⁻⁶/(10x 10⁻²)² NC⁻¹
=5.4 x 10⁶ NC⁻¹ along OB
Therefore, the electric field at mid-point O is 5.4 x 106 NC⁻¹ along OB.
(b).
A test charge of amount 1.5 x 10-9 C is placed at mid-point O.
q= 1.5 x 10⁻9C
Force experienced by the test charge = F (say)
Then F = qE
= 1.5 x 10⁻9 x 5.4 x 106 = 8.1 x 10-3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 x 10-3 N along OA.
See less(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?
(a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other. (b).If two field lines cross eachRead more
- (a).An electrostatic field line is a continuous curve because a charge experiences a continuous force when traced in an electrostatic field. The field line cannot have sudden breaks because the charge moves continuously and does not jump from one point to the other.
- (b).If two field lines cross each other at a point, then electric field intensity will show two directions at that point. This is not possible. Hence, two field lines never cross each other.
See less