Physics Q10 – NEET-UG 2024

In a uniform magnetic field of

0.049

T, a magnetic needle performs

20

complete oscillations in

5

seconds as shown. The moment of inertia of the needle is

9.8 \times 10^{- 6}

kg m

^{2}

. If the magnitude of magnetic moment of the needle is

x \times 10^{- 5}

^{2}

; then the value of 'x' is :

A magnetic needle suspended in a uniform magnetic field, when disturbed from its equilibrium position, performs angular simple harmonic motion.

🥷

Ninja Strategy•Order of Magnitude Estimation

Estimate the order of magnitude of the magnetic moment using approximate values for the given parameters to quickly narrow down the options.

Step 1: Calculate the Period of Oscillation✦ Active

The magnetic needle performs $20$ complete oscillations in $5$ seconds. The period of oscillation $T$ is the time taken for one complete oscillation.

T = \frac{Total time}{Number of oscillations} = \frac{5 s}{20} = 0.25 s

Step 2: Apply the Formula for Period of Magnetic Needle Oscillation○ Expand

The period of oscillation of a magnetic needle with moment of inertia $I$ and magnetic moment $M$ in a uniform magnetic field $B$ is given by:

T = 2 π \sqrt{\frac{I}{M B}}

Squaring both sides and rearranging to solve for $M$ :

T^{2} = 4 π^{2} \frac{I}{M B} ⟹ M = \frac{4 π^{2} I}{T^{2} B}

💡 Teacher's Secret Hint

Remember to square the period $T$ and $2 π$ when rearranging the formula.

Step 3: Substitute Values and Calculate Magnetic Moment○ Expand

Given values are $I = 9.8 \times 10^{- 6}$ kg m $^{2}$ , $B = 0.049$ T, and $T = 0.25$ s. Substitute these into the rearranged formula:

M = \frac{4 π^{2} (9.8 \times 10^{- 6})}{(0.25)^{2} (0.049)}

Simplify the expression:

M = \frac{4 π^{2} \times 9.8 \times 10^{- 6}}{0.0625 \times 0.049} = \frac{4 π^{2} \times 9.8 \times 10^{- 6}}{(1 / 16) \times 0.049}

M = \frac{64 π^{2} \times 9.8 \times 10^{- 6}}{0.049} = \frac{64 π^{2} \times (2 \times 0.049) \times 10^{- 4}}{0.049}

M = 128 π^{2} \times 10^{- 4} {Am}^{2}

The magnetic moment is given as $x \times 10^{- 5}$ Am $^{2}$ . Convert the calculated value to this format:

M = 128 π^{2} \times 10^{- 4} = 128 π^{2} \times 10 \times 10^{- 5} = 1280 π^{2} \times 10^{- 5} {Am}^{2}

Comparing this with $x \times 10^{- 5}$ Am $^{2}$ , we find $x = 1280 π^{2}$ .

💡 Teacher's Secret Hint

Pay close attention to the powers of $10$ and the final required format for $x$ .

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Physics Question 10 – NEET-UG 2024