Question

A conductor of length 50 cm carrying of 5 a is placed perpendicular to a magnetic field of induction 2 x 10-3 t. Find the force on the conductor.

Answer 1

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Answer 2

Answer:

The force on the conductor is 5×10⁻³N.

Explanation:

The force on the conductor present in a magnetic field is given as,

$F=BIL$     (1)

Where,

F=force acting on the conductor

B=magnetic field in which the conductor is present

I=current flowing through the conductor

L=length of the conductor

From the question we have,

The length of the conductor(L)=50cm=0.5m

The current flowing through the conductor(I)=5A

The magnetic field in which the conductor is present(B)=2×10⁻³T

By inserting the variables in equation (1) we get;

$F=2\times 10^-^3\times 5\times 0.5$

$F=5\times 10^-^3N$

Hence,  the force on the conductor is 5×10⁻³N.

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