Chapter 24, Problem 97A

(a)

To determine

The force on a current-carrying wire of a given length kept in a uniform magnetic field for a given inclination with the field.

Answer to Problem 97A

The force on the wire for $\theta =90°$ is $3.1875\text{ N}$

Explanation of Solution

Given:

A wire, 25 cm long and carrying a current of 15 A is kept in a uniform magnetic field of strength 0.85 T at an angle of $90°$ with the field.

Formula Used:

The force acting on a wire carrying current is dependent on the strength of magnetic field, the current, length of the wire and the angle which the wire makes with the magnetic field. It is mathematically expressed as,

  $F=ILB\left(\sin \theta \right)$

Calculation:

For the given strength of magnetic field $B=0.\text{85 T}$ , current $I=15\text{ A}$ and length $L=25\text{ cm}$ , the force at angle $\theta =90°$ is given by,

  $\begin{array}{c}F=15\times 25\times {10}^{-2}\times 0.85\times \left(\sin 90°\right)\\ =3.1875\text{ N}\end{array}$

Conclusion:

Therefore, the force on the current carrying wire is 3.1875 N.

(b)

To determine

The force on a current-carrying wire of a given length kept in a uniform magnetic field for a given inclination with the field.

Answer to Problem 97A

The force on the wire for $\theta =45°$ is $2.253\text{ N}$.

Explanation of Solution

Given:

A wire, 25 cm long and carrying a current of 15 A is kept in a uniform magnetic field of strength 0.85 T at an angle of $45°$ with the field.

Formula Used:

The force acting on a wire carrying current is dependent on the strength of magnetic field, the current, length of the wire and the angle which the wire makes with the magnetic field. It is mathematically expressed as,

  $F=ILB\left(\sin \theta \right)$

Calculation:

For the given strength of magnetic field $B=0.\text{85 T}$ , current $I=15\text{ A}$ and length $L=25\text{ cm}$ , the force at angle $\theta =45°$ is given by,

  $\begin{array}{c}F=15\times 25\times {10}^{-2}\times 0.85\times \left(\sin 45°\right)\\ =2.253\text{ N}\end{array}$

Conclusion:

The force on the current carrying wire for the given inclination is 2.253 N.

(c)

To determine

The force on a current-carrying wire of a given length kept in a uniform magnetic field for a given inclination with the field.

Answer to Problem 97A

The force on the wire for $\theta =0°$ is $0\text{ N}$

Explanation of Solution

Given:

A wire, 25 cm long and carrying a current of 15 A is kept in a uniform magnetic field of strength 0.85 T at an angle of $0°$ with the field.

Formula Used:

The force acting on a wire carrying current is dependent on the strength of magnetic field, the current, length of the wire and the angle which the wire makes with the magnetic field. It is mathematically expressed as,

  $F=ILB\left(\sin \theta \right)$

Calculation:

For the given strength of magnetic field $B=0.\text{85 T}$ , current $I=15\text{ A}$ and length $L=25\text{ cm}$ , the force at angle $\theta =0°$ is given by,

  $\begin{array}{c}F=15\times 25\times {10}^{-2}\times 0.85\times \left(\sin 0°\right)\\ =0\text{ N}\end{array}$

Conclusion:

For $\theta =0°$ , the wire is parallel to the field and hence experiences no force.