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. $2 \sqrt { 2 }$ N m-1
• B. $\frac { 2 } { \sqrt { 2 } }$N m-1
• C. $\frac { \sqrt { 2 } } { 2 }$N m-1
• D. $4 \sqrt { 2 }$ N m-1

Answer 1

Answer: (B)

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

Answer 2

$\therefore \frac { \mathrm { F } } { 1 } = \mathrm { IB } \sin 45 ^ { \circ } = 10 \times 0.2 \times \frac { 1 } { \sqrt { 2 } } = \frac { 2 } { \sqrt { 2 } } \mathrm { Nm } ^ { - 1 }$