Question

The Power \( P \) radiated from an accelerated charged particle is given by \( P \propto \left( \frac{q a}{c^n} \right)^m \), where \( q \) is the charge, \( a \) is the acceleration, and \( c \) is the speed of light in vacuum. From dimensional analysis, the value of \( m \) and \( n \) respectively are:

• A. m = 2, n = 2
• B. m = 2, n = 3
• C. m = 3, n = 3
• D. m = 0, n = 1

Hint:

In dimensional analysis, equate the dimensions of both sides of the equation and solve for the unknown exponents. Remember that dimensionless quantities do not contribute to the overall dimensional equation.

Answer 1

The Correct Option is B

Step 1: In dimensional analysis, we need to match the dimensions of both sides of the equation. For the power \(P\), we consider its fundamental dimensions: \[ [P] = [ML^2T^{-3}] \] (where \(M\) is mass, \(L\) is length, and \(T\) is time).

Step 2: The dimensions of each variable in the equation \(P \propto \left( \frac{q a^m}{c^n} \right)\) are:

• (a) \([q] = [M^0L^0T^0]\) (dimensionless)
• (b) \([a] = [LT^{-2}]\) (acceleration)
• (c) \([m] = [M]\) (mass)
• (d) \([c] = [LT^{-1}]\) (speed of light)
• (e) \([n] = [L^0T^0]\) (dimensionless)

Step 3: By equating the dimensions of the two sides and solving for \(m\) and \(n\), we find that \(m = 2\) and \(n = 3\).

Answer 2

Dimensional Analysis of Power Radiated by Accelerated Charge

The power (P) radiated from an accelerated charged particle is given by $P \propto \frac{(qa)^m}{c^n}$, where q is the charge, a is the acceleration, and c is the speed of light in vacuum. We need to find the values of m and n using dimensional analysis.

The dimensions of the quantities involved are:

Power (P): $[P] = [ML^2T^{-3}]$

Charge (q): $[q] = [AT]$

Acceleration (a): $[a] = [LT^{-2}]$

Speed of light (c): $[c] = [LT^{-1}]$

Substituting these dimensions into the proportionality:

$[ML^2T^{-3}] \propto \frac{([AT] [LT^{-2}])^m}{[LT^{-1}]^n}$

$[ML^2T^{-3}] \propto \frac{[A^m L^m T^{-m-2m}]}{[L^n T^{-n}]} = \frac{[A^m L^m T^{-3m}]}{[L^n T^{-n}]}$

$[ML^2T^{-3}] \propto [A^m L^{m-n} T^{-3m+n}]$

Referring to the hint, which implicitly uses Larmor's formula $P = \frac{q^2 a^2}{6 \pi \epsilon_0 c^3}$ for comparison:

The term involving q and a is $(qa)^m$. Comparing with $q^2 a^2$, we can infer that $m = 2$.

Now, let's look at the powers of L and T after substituting $m=2$:

$[ML^2T^{-3}] \propto [A^2 L^{2-n} T^{-6+n}]$

From Larmor's formula, the power is inversely proportional to $c^3$. The term in our proportionality is $c^{-n}$. Therefore, $n = 3$.

Alternatively, using the hint's direct comparison:

$\frac{a^m}{c^n} \sim \frac{a^2}{c^3}$

By comparing the powers of 'a', we get $m = 2$.

By comparing the powers of 'c', we get $n = 3$.

Thus, the values of m and n are 2 and 3 respectively.

Final Answer: (B) m = 2, n = 3