Question

a wire carrying 1.5 A passes through a 48 mT magnetic field the wire is peremndicular to the field and makes a quarter circle turn radius 21 cm in the dield regeion. Find the magnitude and direction of the magnetic force

Answer 1

Answer:

Explanation:

Wire is in the form of an arc of the size of a quarter of a circle. The distance between its two end

=√2 x 21 cm

= 29.7 cm

= 29.7 x 10⁻² m

Force on the arc in the magnetic field will be same as the straight line meeting its two ends.

Force = BIL

= 48 x10⁻³ x  1.5 x 29.7 x 10⁻²

= 2138.4 x 10⁻⁵ N.

Answer 2

The magnitude and direction of the magnetic force will be:

"2138.4 × 10⁻⁵ N (Same as the straight line)"

Magnetic field

According to the question,

Current, I = 1.5 A

Magnetic field, B = 48 mT or,

                            = 48 × 10⁻³ m

Radius, r = 21 cm

Now,

The distance between two ends will be:

= √2 × 21

= 29.7 cm or,

= 29.7 × 10⁻² m

hence,

The force will be:

→ F = BIL

By substituting the values,

      = 48 × 10⁻³ × 1.5 × 29.7 × 10⁻²

      = 2138.4 × 10⁻⁵ N

Thus the above answer is correct.

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