Question:
A wire of length 0.5 m carrying current 4 A is placed perpendicular to a magnetic field of 0.2 T. The force exerted on the wire is
A wire of length 0.5 m carrying current 4 A is placed perpendicular to a magnetic field of 0.2 T. The force exerted on the wire is
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For a wire perpendicular to a magnetic field:
\[
F=BIL
\]
because \(\sin 90^\circ =1\).
Updated On: Apr 27, 2026
- 0.1 N
- 0.2 N
- 0.4 N
- 0.6 N
- 1.0 N
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The Correct Option is C
Solution and Explanation
Force on a current carrying conductor is:
\[
F=BIL\sin\theta
\]
Since the wire is perpendicular to the field:
\[
\theta=90^\circ,\qquad \sin 90^\circ =1
\]
Given:
\[
B=0.2\text{ T},\quad I=4\text{ A},\quad L=0.5\text{ m}
\]
So,
\[
F=0.2\times 4\times 0.5 = 0.4\text{ N}
\]
Hence, the correct answer is: \[ \boxed{(C)\ 0.4\text{ N}} \]
Hence, the correct answer is: \[ \boxed{(C)\ 0.4\text{ N}} \]
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