First slide

Magnetic field due to a straight current carrying wire

Question

A straight wire current element is carrying current 100 A, as shown in figure. The magnitude of magnetic field at point P which is at perpendicular distance $(\sqrt{3} - 1) m$ from the current element if end A and end B of the element subtend angle 30⁰ and 60⁰ at point P, as shown, is

Moderate

A

$5 \times 10^{- 6} T$
B

$2.5 \times 10^{- 6} T$
C

$2.5 \times 10^{- 5} T$
D

$8 \times 10^{5} T$

Solution

$\begin{array}{l} B = \frac{μ_{0} . i}{4 πa} (\sin θ_{1} + \sin θ_{2}) \\ = \frac{10^{- 7} \times 100}{\sqrt{3} - 1} [\frac{\sqrt{3}}{2} - \frac{1}{2}] = 5 \times 10^{- 6} T \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App