A $3.0 c m$ wire carrying a current of $10 A$ is a placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27 T$ . What is magnetic force on the wire?

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Solution

Given,
$i = 10 A, l = 3 c m = 0.03 m,$

$B = 0.27 T and θ = 90^{\circ}$

Magnetic force on a current carrying conductor is given by:
$\to F = i (\to l \times \to B) = i l B sin θ$

$F = 10 \times 0.03 \times 0.27 \times sin 90^{\circ}$

$F = 8.1 \times 10^{- 2} N$

Direction by force is finding out by Fleming left hand rule.

Final Answer: $8.1 \times 10^{- 2} N$ ; direction of force given by Fleming's left-hand rule

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