The magnetic field intensity due to a thin wire carrying current I in the figure shown is
Physics · Class 12 · CBSE — Moving Charges and Magnetism
The magnetic field intensity due to a thin wire carrying current I in the figure shown is
- \(\frac{\mu_{0} I}{2 \pi R}(\pi-\alpha+\tan \alpha)\)
- \(\frac{\mu_{o} I}{2 \pi R}(\pi-\alpha)\)
- \(\frac{\mu_{o} I}{2 \pi R}(\pi+\alpha)\)
- \(\frac{\mu_{0}}{2 \pi R}(\pi+\alpha-\tan \alpha)\)
Answer
(D) _ 0 2 R ( + - )
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