Question

The poles of a horse-shoe magnet each of pole strength $2\text{Am}$ are at $4\text{cm}$ apart.

$\left(ii\right)$The magnetic induction at the mid point between the poles is $\text{(in}\text{T})$

• A. ${10}^{-3}$
• B. $0.125\times {10}^{-3}$
• C. $0.5\times {10}^{-3}$
• D. $0.25\times {10}^{-3}$

Answer 1

The correct option is A ${10}^{-3}$


As we can at the mid, field due to both $\text{N}$ and $\text{S}$ directed along the same direction,

The net field at the mid-point $\text{P}$ is,

$B_{net}=B_N+B_S$

At the equatorial point $B_H=B_s$

$B_{net}=2B_N$

$=2\times \frac{{\mu }_0}{4\pi }\left(\frac{m}{d^2}\right)$

$=2\times [\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{2}{(2\times {10}^{-2}{)}^2}]$

$[\because d=\frac{4}{2}=2\text{cm}=2\times {10}^{-2}\text{m}]$

$=\frac{2\times {10}^{-7}\times 2}{4\times {10}^{-4}}$

$\therefore B_{net}={10}^{-3}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.