Hc Verma II for Class 11 Science Physics Chapter 35 - Magnetic Field Due To A Current

Share with your friends

Answer:

We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

Page No 248:

Answer:

Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances.If the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Let at any point in between the two wires, B₁ and B₂ be the magnetic field due to wire 1 and wire 2 respectively.

From the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

No. of turns: n₁ = 50 and n₂ = 100
Magnitude of currents: i₁ = i₂ = 2 A
Radii of loops: r₁ = 5 cm and r₂ = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre is given by
$B = \frac{μ_{0} n_{1} i_{1}}{2 r_{1}} + \frac{μ_{0} n_{2} i_{2}}{2 r_{2}} = \frac{4 π \times 10^{- 7} \times 50 \times 2}{2 \times 5 \times 10^{- 2}} + \frac{4 π \times 10^{- 7} \times 100 \times 2}{2 \times 10 \times 10^{- 2}} = 4 π \times 10^{- 4} + 4 π \times 10^{- 4} = 2 \times 4 π \times 10^{- 6} = 8 π \times 10^{- 3} T$

(b) In the opposite sense:
The magnetic field intensity at the centre is given by

$B = \frac{μ_{0} n_{1} i_{1}}{2 r_{1}} - \frac{μ_{0} n_{2} i_{2}}{2 r_{2}} = \frac{4 π \times 10^{- 7} \times 50 \times 2}{2 \times 5 \times 10^{- 2}} - \frac{4 π \times 10^{- 7} \times 100 \times 2}{2 \times 10 \times 10^{- 2}} = 0$

Page No 252:

Answer:

Given:
No. of turns: n₁ = 50 and n₂ = 100
Magnitude of currents: i₁ = i₂ = 2 A
Radii of loops: r₁ = 5 cm and r₂ = 10 cm
(a) In the same sense:

The magnetic field intensity at the centre due to C₁ is given by
$B_{1} = \frac{μ_{0} n_{1} i_{1}}{2 r_{1}} = \frac{4 π \times 10^{- 7} \times 50 \times 2}{2 \times 5 \times 10^{- 2}} = 4 π \times 10^{- 4} T$
(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C₂ is given by
$B_{2} = \frac{μ_{0} n_{2} i_{2}}{2 r_{2}} = \frac{4 π \times 10^{- 7} \times 100 \times 2}{2 \times 10 \times 10^{- 2}} = 4 π \times 10^{- 4} T$

(In the plane of paper in upward direction)

In this case, magnetic fields due to C₁ and C₂ at the centre are along the same direction.
Thus, the net magnetic field is given by
$B_{net} = B_{1} + B_{2} = (4 π \times 10^{- 4}) + (4 π \times 10^{- 4}) = 8 π \times 10^{- 4} T = 25.12 mT$
(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

$B_{net} = B_{1} - B_{2} = (4 π \times 10^{- 4}) - (4 π \times 10^{- 4}) = 0$

Page No 252:

Answer:

Given:
Magnitude of current in the loop, I = 10 A
Radius of the loop, r = 20 cm = 20 × 10⁻² m
Thus, the magnetic field intensity at the centre is given by
$B = \frac{μ_{0} I}{2 r}$
Now,
Velocity of the electron, v = 2 × 10⁶ m/s
Angle between the velocity and the magnetic field intensity, θ = 30°
Thus, the magnetic force on the electron is given by
$F = e v B \sin θ = 1.6 \times 10^{- 19} \times 2 \times 10^{6} \times \frac{μ_{0} i}{2 R} \sin 30 ° = 1.6 \times 10^{- 19} \times 2 \times 10^{6} \times \frac{4 π \times 10^{- 7} \times 10}{2 \times 20 \times 10^{- 2}} \times \frac{1}{2} = 16 π \times 10^{- 19} N$

View NCERT Solutions for all chapters of Class 11