Magnetic Effects Of Current Question 140
Question: A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field $ \overset{\to }{\mathop{B}}, $ at a point having coordinates (x, y) in the z = 0 plane is [IIT-JEE (Screening) 2002]
Options:
A) $ \frac{{\mu_{o}}I,(y\hat{i}-x\hat{j})}{2\pi (x^{2}+y^{2})} $
B) $ \frac{{\mu_{o}}I,(x\hat{i}+y\hat{j})}{2\pi (x^{2}+y^{2})} $
C) $ \frac{{\mu_{o}}I,(x\hat{j}-y\hat{i})}{2\pi (x^{2}+y^{2})} $
D) $ \frac{{\mu_{o}}I,(x\hat{i}-y\hat{j})}{2\pi (x^{2}+y^{2})} $
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Answer:
Correct Answer: A
Solution:
Magnetic field at P is $ \overrightarrow{B} $ , perpendicular to OP in the direction shown in figure. So, $ \overrightarrow{B}=B\sin \theta ,\hat{i}-B\cos \theta ,\hat{j} $ Here $ B=\frac{{\mu_{0}}}{2\pi }\frac{I}{r} $ $ \sin \theta =\frac{y}{r} $ and $ \cos \theta =\frac{x}{r} $ \ $ \overrightarrow{B}=\frac{{\mu_{0}}I}{2\pi }\cdot \frac{1}{r^{2}}(y\hat{i}-x\hat{j})=\frac{{\mu_{0}}I(y\hat{i}-x\hat{j})}{2\pi (x^{2}+y^{2})} $ (as $ r^{2}=x^{2}+y^{2} $ )
BETA
