A straight wire of length $2 m$ carries a current of $10 A$ . If this wire is placed in a uniform magnetic field of $0.15 T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be:

1.5 N

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3 N

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3 \sqrt{2} N

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\frac{3}{\sqrt{2}} N

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Solution

The correct option is C $\frac{3}{\sqrt{2}} N$
Given: $i = 10 A, B = 0.15 T, θ = 45^{\circ}$ and $l = 2 m$
Force on a current carrying wire of finite length $= \to F = l (\to i \times \to B)$
Hence, magnitude of force, $F = B i l s i n θ$
$= 0.15 \times 10 \times 2 \times s i n 45^{\circ}$
$= \frac{3}{\sqrt{2}} N$

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A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of 45∘ with the magnetic field, the applied force on the wire will be:

The correct option is C 3√2NGiven: i=10A,B=0.15 T,θ=45∘ and l=2 mForce on a current carrying wire of finite length =→F=l(→i×→B)Hence, magnitude of force, F=Bilsinθ =0.15×10×2×sin45∘ =3√2N

A straight wire of length $2 m$ carries a current of $10 A$ . If this wire is placed in a uniform magnetic field of $0.15 T$ making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be: