Moving Charges And Magnetism

The correct option is (C): 4B

B₁ = B =
B₂ =
2 × 2πr = 4πR ∴ r = ∵ 2×2πr=4πR ∴ r=

The correct option is (B) : q E + q ( V × B)

Lorentz forece F_L = F_e + F_m = q E + q( v × B)

Let and are the radius of coil 1 & 2.
If and are magnetic induction at their centre, then
; and
Since ; and
therefore
Again if and are resistance of the coil 1 and 2 then
$
and if and are the potential difference across them respectively, then


Therefore, the correct option is (C) : 4.

Frequency Calculation

The formula for frequency is given by:

Where: - is the charge, - is the magnetic field strength, - is the mass of the particle.

Rearranging the formula, we get:

Substituting the given values: - , - , - ,

Simplifying the expression:

Finally, converting to GHz:

Conclusion:

The frequency is .

Magnetic Field Inside and Outside a Wire

For a current-carrying wire, the magnetic field at any point depends on the position relative to the wire. The equations for the magnetic field are different for points inside and outside the wire:

1. For Points Inside the Wire

The magnetic field at a point inside the wire (at a distance from the center of the wire, where , and is the radius of the wire) is given by:

2. For Points Outside the Wire

The magnetic field at a point outside the wire (at a distance from the center of the wire, where ) is given by:

3. Ratio of Magnetic Fields

According to the question, we are comparing the magnetic fields at two points:

Simplifying the expression:

Conclusion:

The ratio of the magnetic fields at the two points is , meaning the magnetic fields are equal.

To solve this question, we need to understand the magnetic field variation inside and outside a thick current-carrying cable. The cable has a radius 'R' and carries a uniformly distributed current 'I'. We want to determine how the magnetic field 'B' varies with the distance 'r' from the axis of the cable.

According to Ampère's law, the magnetic field inside and outside a current-carrying conductor is given by:

1. Inside the cable (r < R):
2. Outside the cable (r ≥ R):

Given these expressions, the graph for the magnetic field as a function of distance 'r' from the axis of the cable will increase linearly with 'r' for and decay as for .

Let's match this to the options provided in the question:

This option shows the correct linear increase within the cable followed by a decrease following outside the cable.

Therefore, the correct option for the variation of the magnetic field with the distance 'r' is the one represented by the fourth option.

Ampere’s circuital law for a long straight wire states that the line integral of the magnetic field B around a closed path is equal to µ₀ times the total current I enclosed by the path. Mathematically, this is expressed as:

$ (R being the radius of the wire), the magnetic field is determined by the fraction of the current enclosed by the circle of radius r. If the current density is uniform, the enclosed current at radius r is . Applying Ampere’s law: \) , the entire current I is enclosed within the loop. By Ampere’s law: \) $ Hence, B decreases with 1/r outside the wire.

Thus, the correct description of the variation of magnetic field is that it is a linearly increasing function of distance r up to the boundary of the wire and then decreases with a 1/r dependence for the outside region.

The problem involves calculating the magnetic field at point P due to a very long conducting wire bent in a semi-circular shape. To solve this, we can utilize the Biot-Savart Law, which relates the magnetic field to a current flowing through a conductor as follows:

Given the configuration, we need to find the magnetic field contribution from the straight wires and the semi-circular part separately, then combine these contributions.

1. Semi-circular part AB: The magnetic field at the center of a current-carrying circular arc is given by:

where for a semi-circle, simplifying to:

This field points out of the page according to the right-hand rule.

1. Straight parts: For a long straight wire segment, the magnetic field contribution at point P from both lines AB and the corresponding part of the other side is zero because the field at the center from infinite parts cancel each other out.

Thus, the net magnetic field is only due to the semi-circular part of the wire. However, since the same calculation applies for a full circle:

For the semi-circle, half of this value applies, needing adjustment for uniformity in integration:

which simplifies to:

This net magnetic field points away from the page.

Hence, the correct option is: pointed away from the page.