JEE Physics 2025 Question with Solution

Given: Total length of the bar magnet is $2l = 20$ units, so $l = 10$ units. The point $P$ is at distance $d = 10$ units from the centre. Relative uncertainty in length measurement is $1\%$.

Find: The relative uncertainty in the magnetic field at point $P$.

From the extracted solution, the magnetic field is taken to vary so that the propagated percentage uncertainty gets multiplied by a factor of $5$.

Using error propagation,

$$\frac{\Delta B}{B} = 5 \frac{\Delta l}{l}$$

Now substitute $\frac{\Delta l}{l} = 1\%$:

$$\frac{\Delta B}{B} = 5 \times 1\% = 5\%$$

Therefore, the uncertainty in the magnetic field is $5\%$. The correct option is C.

Given:

• Total length of the bar magnet, $2l = 20$, hence $l = 10$ units
• Distance of point $P$ from the center, $d = 10$ units
• Relative uncertainty in length measurement = $1\%$

Find: Relative uncertainty in magnetic field at point $P$.

The extracted solution states that for the axial field of a bar magnet,

$$B = \frac{\mu_0}{4\pi} \frac{2M}{1} \left[ \frac{1}{(d-l)^2} - \frac{1}{(d+l)^2} \right]$$

which is simplified there to

$$B \propto \frac{4dl}{(d^2-l^2)^2}$$

So,

$$B = k \frac{dl}{(d^2-l^2)^2}$$

where $k$ is a constant.

The extracted working then applies propagation of errors and concludes that near the given values, the uncertainty gets amplified by a factor of $5$:

$$\frac{\Delta B}{B} = 5 \frac{\Delta l}{l}$$

Since

$$\frac{\Delta l}{l} = 1\%$$

we get

$$\frac{\Delta B}{B} = 5 \times 1\% = 5\%$$

Conclude: The uncertainty in the magnetic field is $5\%$, so the correct option is C.

Note: The first extracted approach mentions a factor of $3$ based on $B \propto \frac{1}{d^3}$, but the solution explicitly marks option C and the detailed extracted working concludes $5\%$. Hence the final answer is taken as C.