The horizontal component of earths magnetic field at a certain place is 3.0 × 10-5 T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of 1.2 A in east to west direction is

Question

The horizontal component of earths magnetic field at a certain place is 3.0 × 10-5 T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of 1.2 A in east to west direction is (A) 3.0 × 10-5 N m-1 (B) 3.2 × 10-5 N m-1 (C) 3.6 × 10-5 N m-1 (D) 3.8 × 10-5 N m-1

Tardigrade Admin · Accepted Answer

Force per unit length f=(F/l)=IB sin θ When the current is flowing from east to west then θ=90°, hence f=IBsin90°=1.2×3×10-5×1 =3.6×10-5 N m-1

Q. The horizontal component of earths magnetic field at a certain place is $3.0 \times 1 0^{- 5} T$ and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of $1.2 A$ in east to west direction is

$3.0 \times 1 0^{- 5} N m^{- 1}$

$3.2 \times 1 0^{- 5} N m^{- 1}$

$3.6 \times 1 0^{- 5} N m^{- 1}$

$3.8 \times 1 0^{- 5} N m^{- 1}$

Force per unit length $f = \frac{F}{l} = I B s in θ$
When the current is flowing from east to west then $θ = 9 0^{\circ}$ , hence
$f = I B s in 9 0^{\circ} = 1.2 \times 3 \times 1 0^{- 5} \times 1$
$= 3.6 \times 1 0^{- 5} N m^{- 1}$

Q. The horizontal component of earths magnetic field at a certain place is 3.0×10−5T and having a direction from the geographic south to geographic north. The force per unit length on a very long straight conductor carrying a steady current of 1.2A in east to west direction is

3.0×10−5Nm−1

3.2×10−5Nm−1

3.6×10−5Nm−1

3.8×10−5Nm−1