Question

A wire 1.0m long is parallel to a magnetic field of 0.10T. What is the force on the wire when it carries a current of 10A.​

Answer 1

The force on the wire is zero

Explanation:

Given:

Length of the wire L = 1 m

Magnetic Field B = 0.1 T

Current in the wire i = 10 A

To find out:

The force on the wire

Solution:

We know that if the angle between the current carrying wire and the magnetic field is θ, then the force on the wire is

F = BiLsinθ

But in this question it is given that θ = 0

Therefore, the force

F = 0

Hope this answer is helpful.

Know More:

Q: Derive an expression for rhe force acting on the current carrying conductor placed in a uniform magnetic field name rhe rule which give the direction of the force write the condition for qhich this force will have maximum and minimum value

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Answer 2

Given:

A wire 1.0m long is parallel to a magnetic field of 0.10 T.

To find:

Force on the wire when it carries a current of 10A?

Calculation:

General expression for force on a wire placed in magnetic field is as follows:

$\therefore \: F = i \times L \times B \times \sin( \theta)$

• i is the current
• L is the length
• B is magnetic field intensity
• $\sin(\theta)$ is the angle between length vector and magnetic field intensity vector.

$\implies \: F = 10 \times 1 \times 0.1 \times \sin( \theta)$

• Now, since the length vector and the magnetic field intensity vector is parallel, value of $\theta$ = 0°.

$\implies \: F = 10 \times 1 \times 0.1 \times \sin( {0}^{ \circ} )$

$\implies \: F = 10 \times 1 \times 0.1 \times 0$

$\implies \: F = 0 \: N$

So, force on the wire will be zero (0) N.