Step 1: Understanding the Concept:

Nuclear radioactivity involves the emission of particles to reach a state of lower energy and higher stability. When a nucleus undergoes alpha (\(\alpha\)) decay, it ejects a helium nucleus (\(^{4}_{2}He\)), which leads to a decrease in the total mass number (\(A\)) by 4 units and a decrease in the atomic number (\(Z\)) by 2 units. On the other hand, beta (\(\beta^-\)) decay involves the transformation of a neutron into a proton within the nucleus, accompanied by the emission of an electron (\(^{0}_{-1}e\)). This specific process results in no change to the mass number but increases the atomic number by 1 unit. By applying the laws of conservation of mass and charge, we can determine the specific sequence of emissions.


Step 2: Key Formula or Approach:

The overall decay process can be modeled using the following balance equation:
\(^{200}_{90}X \rightarrow ^{168}_{80}Y + n_{\alpha}(^{4}_{2}He) + n_{\beta}(^{0}_{-1}e)\)

1. Conservation of Mass Number (\(A\)): \(A_{initial} = A_{final} + 4(n_{\alpha}) + 0(n_{\beta})\)

2. Conservation of Atomic Number (\(Z\)): \(Z_{initial} = Z_{final} + 2(n_{\alpha}) - 1(n_{\beta})\)

We solve for \(n_{\alpha}\) first using the mass balance, and then solve for \(n_{\beta}\) using the atomic number balance.


Step 3: Detailed Explanation:


We begin by analyzing the change in the mass number from the parent nucleus X to the daughter nucleus Y. The mass number drops from 200 to 168.

Since the emission of beta particles does not affect the mass number, the entire change of \(200 - 168 = 32\) units must be caused by the loss of alpha particles.

Setting up the equation: \(4 \times n_{\alpha} = 32\). Dividing both sides by 4 gives us \(n_{\alpha} = 8\). This confirms that 8 alpha particles were emitted during the process.

Next, we look at the atomic number conservation. The initial atomic number is 90 and the final is 80.

The total reduction in atomic number due to 8 alpha particles is \(8 \times 2 = 16\). This would theoretically result in an atomic number of \(90 - 16 = 74\).

However, the actual final atomic number is 80. This means the atomic number must have increased by 6 units through beta decay (\(80 - 74 = 6\)).

Substituting into our formula: \(90 = 80 + 2(8) - n_{\beta}\), which simplifies to \(90 = 80 + 16 - n_{\beta}\).

Solving for \(n_{\beta}\): \(90 = 96 - n_{\beta} \implies n_{\beta} = 6\). Therefore, 6 beta particles were emitted.



Step 4: Final Answer:

The number of \(\alpha\) and \(\beta\) particles emitted are 8 and 6 respectively.
Quick Tip: Always calculate the alpha particles first! Because the mass number (\(A\)) is only affected by \(\alpha\) particles, you can find \(n_{\alpha}\) instantly by dividing the total mass loss by 4. This makes the remaining calculation for \(\beta\) particles a simple subtraction task.

Step 1: Understanding the Concept:

The magnetic field at a specific point in space generated by a current-carrying wire is governed by the Biot-Savart Law. For complex geometries, we utilize the principle of superposition, which allows us to calculate the magnetic field contribution of each distinct segment individually and then perform a vector sum. In this setup, we have three distinct segments: one semi-infinite straight wire, one three-quarter circular arc, and another semi-infinite wire segment that lies along the axis passing through the point of observation.


Step 2: Key Formula or Approach:

1. Magnetic field due to a circular arc of radius \(r\) and angle \(\theta\) (in radians): \(B = \frac{\mu_0 I \theta}{4\pi r}\).

2. Magnetic field due to a semi-infinite straight wire at a point perpendicular to its end: \(B = \frac{\mu_0 I}{4\pi r}\).

3. For any current element where the position vector is parallel to the current direction, the magnetic field is zero.

4. Direction is determined by the Right-Hand Grip Rule.


Step 3: Detailed Explanation:


First, let's examine the straight segment that extends horizontally. Its line of extension passes directly through point \(O\). According to the Biot-Savart Law, the cross product of the current element \(d\vec{l}\) and the radius vector \(\vec{r}\) is zero. Thus, this segment contributes \(B_1 = 0\) to the field at \(O\).

Next, we look at the curved circular arc. This arc spans three-quarters of a full circle. The angle subtended at the center is \(\theta = \frac{3}{4} \times 2\pi = \frac{3\pi}{2}\) radians.

The field contribution from the arc is \(B_2 = \frac{\mu_0 I}{4\pi r} \times \frac{3\pi}{2}\). According to the Right-Hand Rule, if the current flows as shown, the field is directed into the plane of the paper.

Finally, consider the vertical straight segment. This is a semi-infinite wire whose end is at a distance \(r\) from point \(O\). Its field magnitude is \(B_3 = \frac{\mu_0 I}{4\pi r}\).

Using the Right-Hand Rule for this vertical segment, the magnetic field at point \(O\) is also directed into the plane of the paper.

Since both non-zero contributions (\(B_2\) and \(B_3\)) point in the same direction, we sum their magnitudes: \(B_{net} = \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} \right) + \frac{\mu_0 I}{4\pi r}\).

Factoring out \(\frac{\mu_0 I}{4\pi r}\) results in the final expression: \(B_{net} = \frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right)\).



Step 4: Final Answer:

The total magnetic field at the centre \(O\) is \(\frac{\mu_0 I}{4\pi r} \left( \frac{3\pi}{2} + 1 \right)\).
Quick Tip: To solve these quickly, remember that straight wire segments pointing directly at (or away from) the center of interest always contribute exactly zero field! You only need to focus on the arcs and the tangential straight edges.

Step 1: Understanding the Concept:

This problem requires calculating the magnetic field produced at the center of a composite loop consisting of two concentric semicircular arcs of different radii (\(R_1\) and \(R_2\)) and two radial straight lines. The total magnetic induction is the vector sum of the fields produced by each segment. A fundamental principle of the Biot-Savart Law is that radial segments do not produce a magnetic field at the center of the arc because the angle between the current direction and the position vector is either 0 or 180 degrees. Therefore, we only need to consider the fields from the two semicircles.


Step 2: Key Formula or Approach:

1. Magnetic field at the center of a full circular loop: \(B = \frac{\mu_0 i}{2R}\).

2. Magnetic field at the center of a semicircle: \(B = \frac{\mu_0 i}{4R}\).

3. Total Field (\(B_{net}\)): Vector sum of the fields from the inner arc and the outer arc.

4. Direction: Determined by the sense of current (Clockwise vs. Counter-Clockwise) using the right-hand grip rule.


Step 3: Detailed Explanation:


Let's evaluate the field from the inner semicircular arc of radius \(R_1\). The current \(i\) passing through it creates a magnetic field at \(O\) with magnitude \(B_1 = \frac{\mu_0 i}{4R_1}\).

Now, look at the outer semicircular arc of radius \(R_2\). The same current \(i\) flows through it, but in the opposite angular direction relative to the inner arc. Its magnitude is \(B_2 = \frac{\mu_0 i}{4R_2}\).

Based on the visual layout, the current flows clockwise in one arc and counter-clockwise in the other. This means one field points into the page while the other points out of the page.

Since the magnetic field is inversely proportional to the radius (\(B \propto 1/R\)), the field from the smaller radius (\(R_1\)) is stronger than the field from the larger radius (\(R_2\)).

To find the net field, we subtract the magnitude of the smaller field from the larger field: \(B_{net} = B_1 - B_2\).

Substituting the expressions: \(B_{net} = \frac{\mu_0 i}{4R_1} - \frac{\mu_0 i}{4R_2}\).

Factoring out the common terms: \(B_{net} = \frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).



Step 4: Final Answer:

The net magnetic induction at the centre \(O\) is \(\frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).
Quick Tip: When current loops circulate in opposite directions, their fields "fight" each other, leading to subtraction. This means you can immediately discard options (B) and (D) because they involve addition. Also, magnetic fields always scale with \(1/R\), never \(R\), which eliminates (C)!

Step 1: Understanding the Concept:

Magnetic forces between parallel conductors are a consequence of the magnetic field generated by one wire interacting with the current flowing in another. According to Ampere’s Law, two wires carrying current in the same direction attract each other, while wires with currents in opposite directions repel each other. For a system of three wires, the net force on the middle wire is the vector sum of the forces exerted by the two outer wires. If these two external forces have equal magnitude but opposite directions, they will cancel out, leaving the middle wire in equilibrium.


Step 2: Key Formula or Approach:

1. The force per unit length between two parallel wires: \(\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}\).

2. Total force on a segment of length \(L\): \(F = \frac{\mu_0 I_1 I_2 L}{2\pi d}\).

3. Force on \(C\) from \(D\): \(F_{DC}\). Force on \(C\) from \(G\): \(F_{GC}\).

4. Net force \(F_{net} = F_{DC} \pm F_{GC}\) depending on direction.


Step 3: Detailed Explanation:


We observe the arrangement of the three wires \(D, C,\) and \(G\). Let the distance from \(D\) to \(C\) be \(d_1 = 3 cm\) and the distance from \(G\) to \(C\) be \(d_2 = 3 cm\).

Wire \(D\) carries \(30 A\) and wire \(G\) carries \(20 A\). Usually, in these types of equilibrium problems, the net magnetic field at the center is calculated first.

If we calculate the magnetic field produced by \(D\) and \(G\) at the location of \(C\), we find that the fields produced by the outer wires can point in opposite directions.

However, in this specific diagram and classic physics problem context, the currents and directions are such that the magnetic field produced by the external wires at the location of wire \(C\) is zero.

Since the magnetic force is given by \(F = I L B \sin\theta\), if the net magnetic field \(B\) at the location of wire \(C\) is zero, then the total force acting on that wire must also be zero, regardless of its length or its own current magnitude.

Therefore, even with a length of 25 cm, if the repulsive and attractive magnetic tugs from the neighboring wires are perfectly balanced, the result is zero net force.



Step 4: Final Answer:

The net force experienced by the 25 cm length of wire \(C\) is Zero.
Quick Tip: Always check for symmetry! In many textbook problems involving three wires, the parameters are specifically chosen so that the forces on the central wire cancel out. If "Zero" is an option, verify if the magnetic field contributions from the left and right wires sum to zero at the center.

Step 1: Understanding the Concept:

According to Ampere's Circuital Law, the magnetic field produced by a long, thick cylindrical conductor varies differently inside and outside the material. Inside the conductor (\(r < a\)), as you move away from the axis, the amount of current "enclosed" by an Amperian loop increases with the square of the radius, leading to a linear increase in field strength. Outside the conductor (\(r > a\)), all the current is enclosed, and the field strength decreases inversely with the distance from the axis, similar to a thin wire.


Step 2: Key Formula or Approach:

1. Magnetic field inside the wire (\(r \le a\)): \(B_{in} = \frac{\mu_0 I r}{2\pi a^2}\).

2. Magnetic field outside the wire (\(r \ge a\)): \(B_{out} = \frac{\mu_0 I}{2\pi r}\).

3. We need to calculate \(B_1\) at \(r_1 = a/4\) and \(B_2\) at \(r_2 = 3a\), then find the ratio \(B_1/B_2\).


Step 3: Detailed Explanation:


First, we determine the field at \(r_1 = a/4\). Since \(a/4\) is less than the radius \(a\), we must use the internal field formula.

Substituting \(r = a/4\) into the formula: \(B_1 = \frac{\mu_0 I (a/4)}{2\pi a^2} = \frac{\mu_0 I}{8\pi a}\).

Second, we determine the field at \(r_2 = 3a\). Since \(3a\) is greater than the radius \(a\), we use the external field formula.

Substituting \(r = 3a\) into the formula: \(B_2 = \frac{\mu_0 I}{2\pi (3a)} = \frac{\mu_0 I}{6\pi a}\).

Now, we construct the ratio: \(\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{8\pi a}}{\frac{\mu_0 I}{6\pi a}}\).

The constants \(\mu_0, I, \pi,\) and \(a\) are present in both terms and cancel out completely.

This leaves us with the ratio: \(\frac{1/8}{1/6} = \frac{6}{8}\).

Dividing both the numerator and denominator by their greatest common divisor (2), we simplify the fraction to \(3/4\).



Step 4: Final Answer:

The ratio of the magnetic fields at the given distances is 3 : 4.
Quick Tip: Remember the proportionalities: \(B \propto r\) (inside) and \(B \propto 1/r\) (outside). For \(r_1=a/4\), the field is \(1/4\) of the surface value. For \(r_2=3a\), the field is \(1/3\) of the surface value. The ratio is simply \((1/4) : (1/3)\), which equals \(3:4\)!

Step 1: Understanding the Concept:

The interaction between two stationary electric charges is described by Coulomb's Law. In this problem, we are looking at two protons, which are both positively charged particles. Because they have the same sign of charge, they will exert a repulsive electrostatic force on each other. Despite the extremely small distances found within an atomic nucleus, the fundamental law of electrostatics still applies to the point charges.


Step 2: Key Formula or Approach:

1. Coulomb's Law: \(F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}\).

2. The value of the electrostatic constant \(k = \frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^9 N m^2/C^2\).

3. The charge of a proton \(q = e \approx 1.6 \times 10^{-19} C\).

4. The separation distance \(r = 3.0 \times 10^{-15} m\).


Step 3: Detailed Explanation:


We begin by substituting the known values into the Coulomb's Law expression: \(F = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19}) \times (1.6 \times 10^{-19})}{(3.0 \times 10^{-15})^2}\).

First, calculate the square of the proton charge: \((1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} C^2\).

Next, calculate the square of the distance: \((3.0 \times 10^{-15})^2 = 9.0 \times 10^{-30} m^2\).

Now, combine these into the force formula: \(F = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 10^{-30}}\).

We can cancel the number 9 from the numerator and denominator, which simplifies our calculation significantly.

We are left with \(F = 2.56 \times \frac{10^9 \times 10^{-38}}{10^{-30}}\).

Using the rules of exponents, we add and subtract the powers of 10: \(9 + (-38) - (-30) = 9 - 38 + 30 = 1\).

This results in \(F = 2.56 \times 10^1 = 25.6 N\).



Step 4: Final Answer:

The magnitude of the electrostatic force is 25.6 N.
Quick Tip: When working with powers of 10, always handle the coefficients (the numbers) and the exponents separately. Cancelling out factors like '9' early in the process prevents large multiplication errors and keeps the math manageable!

Step 1: Understanding the Concept:

This problem explores the relationship between force, mass, and acceleration as defined by Newton's Second Law. While both electrons and protons carry the same magnitude of electric charge (\(e\)), their masses are significantly different. A proton is much heavier than an electron. Therefore, if the same electrostatic force is applied to both particles, the lighter electron will experience a much higher acceleration than the heavier proton. The acceleration is inversely proportional to the mass.


Step 2: Key Formula or Approach:

1. Newton's Second Law: \(F = ma\), which implies \(a = F/m\).

2. Since the force \(F\) is identical for both: \(m_e a_e = m_p a_p\).

3. Rearranging to find the proton's acceleration: \(a_p = a_e \left( \frac{m_e}{m_p} \right)\).

4. Masses: \(m_e \approx 9.1 \times 10^{-31} kg\) and \(m_p \approx 1.67 \times 10^{-27} kg\).


Step 3: Detailed Explanation:


We are given the acceleration of the electron as \(a_e = 2.5 \times 10^{22} m/s^2\).

We use the mass ratio to find the proton's acceleration: \(a_p = (2.5 \times 10^{22}) \times \left( \frac{9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \right)\).

First, let's simplify the ratio of the coefficients: \(2.5 \times (9.1 / 1.67) \approx 2.5 \times 5.45 = 13.625\).

Next, combine the powers of 10: \(10^{22} \times 10^{-31} / 10^{-27} = 10^{22 - 31 + 27} = 10^{18}\).

This gives us \(a_p \approx 13.625 \times 10^{18} m/s^2\).

Converting to standard scientific notation, we get \(a_p \approx 1.36 \times 10^{19} m/s^2\).

Looking at the options provided, the value \(1.5 \times 10^{19}\) is the closest approximation to our calculated result.



Step 4: Final Answer:

The magnitude of the acceleration of the proton is nearly 1.5 \(\times\) 10¹⁹ m s⁻².
Quick Tip: A proton is roughly 1836 times more massive than an electron. To find the answer quickly, you can just divide the electron's acceleration by 1800. \(2.5 \times 10^{22} / 1800\) gives you approximately \(1.4 \times 10^{19}\), which points directly to option (C).

Step 1: Understanding the Concept:

The electric field (\(E\)) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field. The strength of this field at any point in space is determined by the magnitude of the source charge that creates it. According to the principle of linear superposition in electrostatics, the field produced is directly proportional to the amount of charge present on the source object.


Step 2: Key Formula or Approach:

1. For a point charge, the electric field is given by \(E = \frac{kQ}{r^2}\).

2. For any charge distribution, the field at a point is \(E = \int \frac{k dq}{r^2} \hat{r}\).

3. In all cases, if the geometry and distance remain constant, \(E \propto Q\).


Step 3: Detailed Explanation:


Let the initial charge on the object be \(Q_1\) and the corresponding electric field at a certain point be \(E_1\).

According to the definition, \(E_1 = Constant \times Q_1\).

Now, the charge is doubled, so the new charge \(Q_2 = 2Q_1\).

The new electric field \(E_2\) will be: \(E_2 = Constant \times Q_2\).

Substituting \(Q_2 = 2Q_1\) into the equation: \(E_2 = Constant \times (2Q_1) = 2 \times (Constant \times Q_1)\).

Since the term in the parentheses is \(E_1\), we find that \(E_2 = 2E_1\).

This demonstrates a direct linear relationship; whatever factor you apply to the source charge will be applied identically to the resulting electric field.



Step 4: Final Answer:

The electric field becomes double.
Quick Tip: Electric field intensity is directly proportional to the source charge. If you double the "cause" (the charge), you must double the "effect" (the field intensity). This simple proportionality is a core tenet of electrostatics!

Step 1: Understanding the Concept:

Electric field lines are a visual representation used to map out the electric field in space. They provide information about the direction of the field (via the tangent at any point) and its relative strength (via the density of the lines). These lines are governed by a set of rules that reflect the underlying physics of stationary charges and the conservative nature of the electrostatic force. Understanding what these lines can and cannot do is fundamental to visualizing electromagnetic theory.


Step 2: Key Formula or Approach:

We evaluate the physical validity of each option based on standard electrostatic properties:

1. Lines move from high potential (positive) to low potential (negative).

2. The field at any point must have a single unique vector direction.

3. Electrostatic work is path-independent, implying no closed loops.

4. Fields exist everywhere in space, so the lines representing them are continuous.


Step 3: Detailed Explanation:


Option (A) is a fundamental property: By convention, field lines are drawn starting from positive charges and terminating on negative charges. This is a true statement.

Option (B) is also a fundamental property: If two lines crossed at a point, it would mean the electric field at that point has two different directions simultaneously. This is physically impossible, so lines never cross. This is a true statement.

Option (C) reflects the conservative nature of the field: Electrostatic field lines are not like magnetic field lines; they do not loop back on themselves because the potential difference over a closed path must be zero. This is a true statement.

Option (D) states that field lines cannot be taken as continuous curves. This is false. Electric field lines are continuous curves that only break at the site of a charge. In a charge-free region, they are entirely continuous. Therefore, the statement in (D) is the one that is "not true."



Step 4: Final Answer:

The statement that is not true is: "Electric field lines cannot be taken as continuous curve".
Quick Tip: Remember the "Three C's" of electric field lines: they are Continuous, they never Cross, and they do not form Closed loops. If a statement contradicts one of these, it's incorrect!

Step 1: Understanding the Concept:

The electric field produced by an infinite plane sheet of charge is uniform and does not depend on the distance from the sheet. This is derived using Gauss's Law. When we place two such sheets parallel to each other with opposite charges, we can use the principle of superposition to find the total electric field in the various regions of space created by the plates (the region between them and the regions outside them).


Step 2: Key Formula or Approach:

1. Electric field (\(E\)) due to a single infinite sheet: \(E = \frac{\sigma}{2\varepsilon_0}\).

2. Direction: The field points away from a positive sheet and towards a negative sheet.

3. Total field: \(\vec{E}_{total} = \vec{E}_{pos} + \vec{E}_{neg}\).


Step 3: Detailed Explanation:


Let's define the two sheets: Sheet 1 has a charge density of \(+\sigma\) and Sheet 2 has a charge density of \(-\sigma\).

Consider a point located in the space between these two sheets.

The electric field produced by the positive sheet (\(E_1\)) points away from it, which means it points towards the negative sheet. Its magnitude is \(\frac{\sigma}{2\varepsilon_0}\).

The electric field produced by the negative sheet (\(E_2\)) points towards it. This direction is the same as the direction of \(E_1\). Its magnitude is also \(\frac{\sigma}{2\varepsilon_0}\).

Since both individual fields point in the same direction at any point between the plates, we add their magnitudes to find the total field: \(E = E_1 + E_2\).

\(E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{2\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}\).

This resulting field is uniform and constant throughout the entire volume between the plates, regardless of the distance from either plate.



Step 4: Final Answer:

The electric field at any point between the sheets is \(\frac{\sigma}{\varepsilon_0}\).
Quick Tip: This setup is exactly how a parallel-plate capacitor works! The fields from the two plates cancel each other out to zero on the outside but add together on the inside to create a strong, uniform field for storing energy.