Question

A uniform magnetic field of 0.20 × 10−3 T exists in the space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.

Answer 1

Explanation:

Given data in the question  

space Magnetic field, $B=0.20 \times 10^{-3} T$

shifted Distance, ∆r= 50 cm  

We know that

$\begin{aligned}&B=-\frac{d V}{d l}\\&d V=\int^{r_{2}}-\vec{B} \cdot d l\end{aligned}$

Because the magnetic field is uniform it can emerge from the sign of integration.

$\Delta V=-B \cdot(\Delta r)$

∆V is the potential charge.

∴ potential changes

$\begin{aligned}&=-0.2 \times 10^{-3} \times 0.5\\&=-0.1 \times 10^{-3} T-m\end{aligned}$

Here the negative sign indicates a decrease in potential.

Answer 2

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$- 0.1 × {10}^{ - 3} tesla$