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

Question

The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of 45° with the direction of a uniform magnetic field of 0.20 T is (A) 2 √2 N m-1 (B) (2/√2) N m-1 (C) (√2/2)N m-1 (D) 4 √2N m-1

Tardigrade Admin · Accepted Answer

I=10 A, θ=45°, B=0.2 T ∴ F=IlB sin θ ∴ (F/l)=IB sin 45° =10×0.2 × (1/√2)=(2/√2) N m-1

Q. The magnetic force per unit length on a wire carrying a current of $10 A$ and making an angle of $4 5^{\circ}$ with the direction of a uniform magnetic field of $0.20 T$ is

$22 N m^{- 1}$

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

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

$42 N m^{- 1}$

$I = 10 A$ , $θ = 4 5^{\circ}$ , $B = 0.2 T$
$∴ F = I lB s in θ$
$∴ \frac{F}{l} = I B s in 4 5^{\circ} = 10 \times 0.2 \times \frac{1}{2} = \frac{2}{2} N m^{- 1}$

Q. The magnetic force per unit length on a wire carrying a current of 10A and making an angle of 45∘ with the direction of a uniform magnetic field of 0.20T is

22​Nm−1

2​2​Nm−1

22​​Nm−1

42​Nm−1

I=10A, θ=45∘, B=0.2T ∴F=IlBsinθ ∴lF​=IBsin45∘=10×0.2×2​1​=2​2​Nm−1

Q. The magnetic force per unit length on a wire carrying a current of $10 A$ and making an angle of $4 5^{\circ}$ with the direction of a uniform magnetic field of $0.20 T$ is

$22 N m^{- 1}$

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

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

$42 N m^{- 1}$

$I = 10 A$ , $θ = 4 5^{\circ}$ , $B = 0.2 T$
$∴ F = I lB s in θ$
$∴ \frac{F}{l} = I B s in 4 5^{\circ} = 10 \times 0.2 \times \frac{1}{2} = \frac{2}{2} N m^{- 1}$