The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of $45^{0}$ with the direction of a uniform magnetic field of 0.20 T is

2 \sqrt{2} N m^{- 1}

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\frac{2}{\sqrt{2}} N m^{- 1}

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\frac{\sqrt{2}}{2} N m^{- 1}

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4 \sqrt{2} N m^{- 1}

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Solution

The correct option is B $\frac{2}{\sqrt{2}} N m^{- 1}$
$I = 10 A, θ = 45^{o}, B = 0.2 T$
$∴ F = I l B s i n θ$
$∴ \frac{F}{l} = I B s i n 45^{o} = 10 \times 0.2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} N m^{- 1}$

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The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of 450 with the direction of a uniform magnetic field of 0.20 T is

The correct option is B 2√2N m−1I=10A,θ=45o,B=0.2T∴F=IlBsinθ∴Fl=IBsin45o=10×0.2×1√2=2√2Nm−1

The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of $45^{0}$ with the direction of a uniform magnetic field of 0.20 T is

The correct option is B $\frac{2}{\sqrt{2}} N m^{- 1}$
$I = 10 A, θ = 45^{o}, B = 0.2 T$
$∴ F = I l B s i n θ$
$∴ \frac{F}{l} = I B s i n 45^{o} = 10 \times 0.2 \times \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} N m^{- 1}$