A current of 2 A flowing in a wire of length 50 cm- If this wire is lying in a uniform magnetic field of 5- 10-4 N-A-m making an angle of 60o with the field- then the force acting on the wire will be -

The correct option is A 4.33× 10^ 4 N Force acting on the wire , F⃗ = i l⃗×B⃗=ilB sinθ= 2 × ( 50 × 10^ 2) × ( 5 × 10 ^ 4 ) ×sin(60^o)= 4.33 × 10^ 4 N ...

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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

A current of ...

Question

A current of $2 A$ flowing in a wire of length $50 c m$ . If this wire is lying in a uniform magnetic field of $5 \times 10^{- 4} N / A - m$ making an angle of $60^{o}$ with the field, then the force acting on the wire will be :

4.33 \times 10^{- 4} N

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Solution

The correct option is A $4.33 \times 10^{- 4} N$
Force acting on the wire ,
$\to F = i \to l \times \to B = i l B sin θ = 2 \times (50 \times 10^{- 2}) \times (5 \times 10^{- 4}) \times sin (60^{o}) = 4.33 \times 10^{- 4} N$

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A current of 2A flowing in a wire of length 50cm. If this wire is lying in a uniform magnetic field of 5×10−4N/A−m making an angle of 60o with the field, then the force acting on the wire will be :

The correct option is A 4.33×10−4NForce acting on the wire ,→F=i→l×→B=ilBsinθ=2×(50×10−2)×(5×10−4)×sin(60o)=4.33×10−4N

A current of $2 A$ flowing in a wire of length $50 c m$ . If this wire is lying in a uniform magnetic field of $5 \times 10^{- 4} N / A - m$ making an angle of $60^{o}$ with the field, then the force acting on the wire will be :

The correct option is A $4.33 \times 10^{- 4} N$
Force acting on the wire ,
$\to F = i \to l \times \to B = i l B sin θ = 2 \times (50 \times 10^{- 2}) \times (5 \times 10^{- 4}) \times sin (60^{o}) = 4.33 \times 10^{- 4} N$