A magnetic field goes to the left of the screen as shown- While the constant current travels at an angle - to the upper right corner- Determine the force per length of wire on this wire-

The correct option is B IBsinθ Magnetic force on current carrying wire #160; #160; #160; #160; F⃗ = l (I⃗×B⃗) #160; #160; #160; ...

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Standard XII

Physics

Coulomb's Law - Mathematical Form

A magnetic fi...

Question

A magnetic field goes to the left of the screen as shown. While the constant current travels at an angle $θ$ to the upper right corner. Determine the force per length of wire on this wire.

\frac{I B}{c o s θ}

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\frac{I B}{s i n θ}

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I B s i n θ

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I B c o s θ

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Solution

The correct option is B $I B s i n θ$
Magnetic force on current carrying wire $\to F = l (\to I \times \to B)$ where $l$ is the length of the wire
$⟹$ Force per unit length $\frac{F}{l} = I B$ $s i n ϕ$
From figure, angle between $\to I$ and $\to B$ $ϕ = 180^{o} - θ$
$∴$ $\frac{F}{l} = I B$ $s i n (180^{o} - θ) = I B$ $s i n θ$

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A magnetic field goes to the left of the screen as shown. While the constant current travels at an angle θ to the upper right corner. Determine the force per length of wire on this wire.

The correct option is B IBsinθMagnetic force on current carrying wire →F=l(→I×→B) where l is the length of the wire⟹ Force per unit length Fl=IB sinϕ From figure, angle between →I and →B ϕ=180o−θ∴ Fl=IB sin(180o−θ)=IB sinθ

A magnetic field goes to the left of the screen as shown. While the constant current travels at an angle $θ$ to the upper right corner. Determine the force per length of wire on this wire.