An 8 cm long wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. If the magnetic field inside the solenoid is 0.3 T, then magnetic force on the wire is

0.14N

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0.24N

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0.34N

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0.44N

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Solution

The correct option is C 0.24N
$F = I l B s i n θ$
Here, $θ = 90^{o}, I = 10 A,$
$l = 8 c m = 8 \times 10^{- 2} m, B = 0.3 T$
$∴ F = 10 \times 8 \times 10^{- 2} \times 0.3 \times s i n 90^{o} = 0.24 N$

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