Question

The straight wire AB carries a current $I$.The ends of the wire subtend angles ${\theta }_1$ and ${\theta }_2$ at the point P as shown in figure.The magnetic field at the point P is

• A. $\frac{{\mu }_{o}I}{4\pi a}(sin{\theta }_1-sin{\theta }_2)$
• B. $\frac{{\mu }_{o}I}{4\pi a}(sin{\theta }_1+sin{\theta }_2)$
• C. $\frac{{\mu }_{o}I}{4\pi a}(cos{\theta }_1-cos{\theta }_2)$
• D. $\frac{{\mu }_{o}I}{4\pi a}(cos{\theta }_1+cos{\theta }_2)$

Answer 1

The correct option is A $\frac{{\mu }_{o}I}{4\pi a}(sin{\theta }_1-sin{\theta }_2)$

Given,

Current, I along A to B, base a,

Two indicators ${\theta }_1\text{and}{\theta }_2$

Therefore total magnetic field,

$\overrightarrow{B_{net}}=\overrightarrow{B_A+}\overrightarrow{B_B}$

$=\frac{{\mu }_{0}I}{4\pi a}\sin {\theta }_1-\frac{{\mu }_{0}I}{4\pi a}\sin {\theta }_2$

$\overrightarrow{B_{net}}=\frac{{\mu }_{0}I}{4\pi a}\left(\sin {\theta }_1-\sin {\theta }_2\right)$