Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

If a charged particle (charge q) is moving in a circle of radius R at a uniform speed v, then the value of its associated magnetic moment \mu will be:

\frac{qvR}{2}

qvR^2

\frac{qvR^2}{2}

qvR

Step-by-Step Solution

Current Equivalent: A charged particle $q$ moving in a circle constitutes a current $I$ . The current is the rate of flow of charge: $I = q/T$ , where $T$ is the time period of revolution .
Time Period: For a particle moving with speed $v$ in a circle of radius $R$ , the time period is $T = \frac{2\pi R}{v}$ . Substituting this into the current equation: $I = \frac{q}{2\pi R / v} = \frac{qv}{2\pi R}$
Magnetic Moment Formula: The magnetic dipole moment $\mu$ associated with a current loop is given by $\mu = I \times A$ , where $A$ is the area of the loop . $A = \pi R^2$
Calculation: Substituting the expressions for $I$ and $A$ : $\mu = \left( \frac{qv}{2\pi R} \right) (\pi R^2)$ $\mu = \frac{qvR}{2}$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started