Sol. Here, l=3.0cm = 3X10-2m; I=10A; B=0.27T; θ=900; F=?
F = B I l sinθ = 0.27 X 10 X (3X10-2) X sin 900 = 0.27 X 10X 10-2 X 1
=8.1X10-2N
The direction of force is perpendicular to the direction of current as well as of magnetic field.
Sol. Here, l = 80cm= 0.80m; N= 5X400 = 2000; i= 8.0A; D=1.8cm
Magnitude of magnetic field induction at a point well inside the solenoid is
Sol. Here, l=10cm=0.10m; n=20; I=12A; α=300; B=0.80T; τ=?
Area, A=l X l = 0.10 X 0.10 = (0.1)2 m2
Τ=n B I A sinα = 20 X 0.80 X 12 X (0.1)2X sin 300
=0.96Nm
Q.10.Two moving coil metres and have the following particulars:
R1 = 10 Ω, N1 = 30, A1 = 3.6 x 10-3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42, A2 = 1.8 x 10-3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
Sol. For meter M1, R1 = 10 Ω, N1 = 30, A1 = 3.6 x 10-3 m2, B1 = 0.25 T, k1=k
For meter M2, R2 = 14 Ω, N2 = 42, A2 = 1.8 x 10-3 m2, B2 = 0.50 T, k2=k
As, current sensitivity I2 = NB A/k
Sol. Here, B = 6.5X10-4 T, v = 4.8X106 m/s e = 1.6X10-19 C, θ=900, m = 9.1X10-31 kg, r=?
- Force on the moving electron due to magnetic field will be, F = e v sinθ.
The direction of this force is perpendicular to as well as therefore this force will only change the direction of motion of the electron without affecting its velocity i.e., this force will provide the centripetal force to the moving electron and hence, the electron will move on the circular path. If r is the radius of circular path traced by electron, then
evB sin 900 = mv2/r
Q.13. (a) A circular coil of 30 turns and radius 8.0cm carrying a current of 6.0A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0T. The field lines make on angle of 600 with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
- Would your answer change if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered).
Since the torque on the planar loop does not depend upon the shape, in case the area of the loop is the same, the torque will remain unchanged.
Sol. For coil X,
r=16cm = 0.16m, n=20, I=16A
Magnetic field induction at the centre of the coil X is given by
We may have, I = 10A; n=800. The core length may be 50cm having 400 turns and area of cross-section is 5X 10-3m2 which is five times given value.
Q.16. For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by B =
- Show that this reduces to the familiar result for field at the centre of the coil.
- Consider two parallel coaxial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separately by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R and is given by B = 0.72
[Such as arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Which is the same as the standard result.
(b)Here, a =R, n=N. Let d be the distance of point P from the mid-point O lying on the axis of the two circular coils C and D, where d <<R.
We know that magnetic field at a point on the axis of the circular coil of radius a, having n turns and carrying current I, which is at a distance x from the centre of the coil is given by
It acts along PO2
Total magnetic field at P due to current through the coils is given by B = B1 +B2
Sol. Here, r1 = 0.25m; r2 = 0.26m; N=3500; I = 11A
- Outside the toroid, the magnetic field is zero.
- Inside the core of the toroid, the magnetic field induction is, B = µ0nI, where n is the number of turns per unit length of toroid = N/I
- In the empty space surrounded by toroid, the magnetic field is zero.
Q.18. Answer the following questions:
- A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
- A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
- An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be setup to prevent the electron from deflecting from its straight line path.
Sol. (a) The magnetic field is in constant direction from East to West. According to the question, a charged particle travels undeflected along a straight path with constant speed. It is only possible, if the magnetic force experienced by the charged particle is zero.
The magnitude of magnetic force on a moving charged particle in a magnetic field is given by F = qv B sin θ. (where θ is the angle between v and B). Here F = 0, if and only if sin θ = 0 (as v ≠ 0, q≠Q, B≠0). This indicates the angle between the velocity and magnetic field is 0° or 180°. Thus, the charged particle moves parallel or anti-parallel to the magnetic field B.
(b)Yes, the final speed be equal to its initial speed as the magnetic force acting on the charged particle only changes the direction of velocity of charged particle but cannot change the magnitude of velocity of charged particle.
- As, the electric field is from North to South, that means the plate in North is positive and in South is negative. Thus, the electrons (negatively charged) attract towards the positive plate that means move towards North. If we want that there should be no deflection in the path of electron, then the magnetic force should be in South direction.
By = — e ( X ), the direction of velocity is West to East, the direction of force is towards South, by using the Fleming’s left hand rule, the direction of magnetic field (B) is perpendicularly inwards to the plane of paper.
Force on electron due to transverse magnetic field is = Bev, which is perpendicular to as well as . It will provide the required centripetal force to the electron for its circular motion. Therefore, the trajectory of electron in magnetic field is circular. Its radius r can be given by
When electron makes an angle θ with the direction of magnetic field, its component velocity perpendicular to the field,
Sol. Here, B= 0.75T, E=9.0 x 105 V m-1, V=15kV =15000V, e/m=?
Let e be the charge and m be the mass of the particles. Let v be the velocity acquired by the particles, when accelerated
hence particles are deuterons ions. But the above value of e/m is also for He++ and Li+++. [As e/m = 2e/2m = 3 e/3 m], so the particles can be He++ or Li+++ also.
Sol. Here, l = 0.45m; m=60g = 60X10-3 kg, I=5.0A
- Tension in the wire is zero if force on the wire carrying current due to magnetic field is equal and opposite to the weight of wire i.e., BIl = mg or mg/Il = (60X10-3)X9.8(5.0X0.45) = 0.26T
The force due to magnetic field will be upwards if the direction of field is horizontal and normal to the conductor.
- When the direction of current is reversed, BIl and mg will act vertically downwards, the effective tension in the wires.
T = BIl + mg = 0.26X5.0X0.45 +(60X10-3)X9.8 = 1.176N
Sol. Given that,
Strength of the magnetic field is, B = 1.5 T
Radius of the cylindrical region, r = 10 cm = 0.1 m
Current in the wire passing through the cylindrical region, I = 7 A
(a) Here the wire is intersecting the axis, so the length of the wire is the diameter
of the cylindrical region. Thus, 𝑙 = 2r = 0.2 m
The angle between the magnetic field and current, θ = 90°
We know that expression for magnetic force acting on the wire is,
F = BII sin θ = 1.5 × 7 × 0.2 × sin 90° = 2.1 N
Hence, a force of magnitude 2.1 N acts on the wire in a vertically downward direction.
(b) The new length of the wire after turning it to the Northeast-Northwest direction can be written as:
L1 = 𝑙/sinθ
Now angle between the magnetic field and current, θ = 45°
Force on the wire,
= BI𝑙1 sin θ
= 1.5 × 7 × 0.2
= 2.1 N
Hence, a force of 2.1 N acts vertically downward on the wire and force is
independent of angle θ because 𝑙 sinθ is fixed.
(c) The wire is lowered from the axis by distance, d = 6.0 cm
Assume that 𝑙2 be the new length of the wire.
= 4(10 + 6) = 4(16)
∴ 𝑙2 = 8 × 2 = 16 cm = 0.16 m
The magnetic force exerted on the wire,
f2 = BI𝑙2
= 1.5 × 7 × 0.16
= 1.68 N
Hence, a force of magnitude 1.68 N acts a vertically downward direction on the wire
Sol. Magnetic field strength, B = 3000 G = 3000 × 10-4 T = 0.3 T
Length of the rectangular loop, l = 10 cm
Width of the rectangular loop, b = 5 cm
Area of the loop,
A = l × b = 10 × 5 = 50 cm2 = 50 × 10-4 m2
Current in the loop, I = 12 A
Now, taking the anti-clockwise direction of the current as positive and vice-versa:
(a) Torque, = I
From the given figure, it can be observed that A is normal to the y-z plane and B is directed along the z-axis.
؞ t = 12 × (50×10-4)
= – 1.8 × 10-2 Nm
The torque 1.8 x 10-2 is N m along the negative y-direction. The force on the loop is zero because the angle between A and B is zero.
(b) This case is similar to case (a). Hence, the answer is the same as (a).
(c) Torque = I
From the given figure, it can be observed that A is normal to the x-z plane and B is directed along the z-axis.
؞ t = 12 × (50×10-4)
= – 1.8 × 10-2 Nm
The torque 1.8 x 10-2 is N m along the negative x direction and the force is zero.
(d) Magnitude of torque is given as:
t = IAB
=12×50×10-4×0.3
= 1.8×10-2 Nm
Torque is 1.8 x 10-2 N m at an angle of 240° with positive x direction. The force is zero.
(e) Torque = I
= (50×10-4×12)
= 0
Hence, the torque is zero. The force is also zero.
(f) Torque = I
= (50×10-4×12)
= 0
Hence, the torque is zero. The force is also zero.
In case (e), the direction of I and is the same and the angle between them is zero. If displaced, they come back to an equilibrium. Hence, its equilibrium is stable.
Whereas, in case (f), the direction of I and is opposite. The angle between them is 180°. If disturbed, it does not come back to its original position. Hence, its equilibrium is unstable.
Q.25. A circular coil of 20 turns and radius 10cm is placed in a uniform magnetic field of 0.10T normal to the plane of the coil. If the current in the coil is 5.0A, what is the
- Total torque on the coil
- Total force on the coil
- Average force on each electron in the coil due to magnetic field? The coil is made of copper wire of cross-sectional area 10-5m2 and the free electron density in copper is given to be about 1029 m-3.
Sol. Here, n=20, r=10cm = 0.10m, B=0.10T, α=00, I=5.0A
Area of the coil, A = =22/7 (o.10)2 m2
- Now, τ = nIBA sin α = nIBA sin00 = 0
- Total force on a planar current loop in a magnetic field is always zero.
- Given, N=1029m-3, A=10-5m2
Force on an electron of charge e, moving with drift velocity vd in the magnetic field is given by
Sol. Here for solenoid, l = 60cm=0.60m; N = 3X300=900:
For wire, l1= 2.0cm=0.02m; m=2.5g = 2.5X10-3kg; I1=6.0A
Let current I be passed through the solenoid windings, the magnetic field produced inside the solenoid due to current is
Q.27. A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 3mA. How will you convert the meter into a voltmeter of range 0 to 18V?
This low resistance called shunt resistance is to be connected in parallel with the galvanometer.