A straight wire carrying a current 10, A is bent into a semicircular arc of radius 5, cm. The magnitude of magnetic field at the

Q: A straight wire carrying a current $10\, A$ is bent into a semicircular arc of radius $5\, cm.$ The magnitude of magnetic field at the center is

c (c) $B = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{\pi i}}{r} \Rightarrow B = {10^{ - 7}} \times \frac{{\pi \times 10}}{{5 \times {{10}^{ - 2}}}} = 6.28 \times {10^{ - 5}}\,T$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A straight wire carrying a current $10\, A$ is bent into a semicircular arc of radius $5\, cm.$ The magnitude of magnetic field at the center is

A$1.5 \times {10^{ - 5}}\,T$
B$3.14 \times {10^{ - 5}}\,T$
C$6.28 \times {10^{ - 5}}\,T$
D$19.6 \times {10^{ - 5}}\,T$

Medium

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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