Question

0. A wire of length 5 cm. is placed inside the
solenoid near its centre such that its makes an
angle of 30° with the axis of solenoid. The wire
carries a current of 5A and the magnetic field
due to solenoid is 2.5 x 10-2T. Magnetic force on
the wire is :
(1) 3.12 * 10-N (2) 31.2 x 10 N
(3) 312 x 10-N (4) 0.312 x 10-4N​

Answer 1

Answer:

MAGNETIC FORCE ( F ) = 31.25 × 10-⁴ N

OPTION 4

Explanation:

★ MAGNETIC FORCE ★

★ GIVEN ;

» Current in Wire ( i ) = 5 A

» Length of Wire ( L ) = 5 cm

» Magnetic Field ( B ) = 2.5 × 10-² T

» Angle Made by Wire with Magnetic Field ( ∅ ) = 30°

★ MAGNETIC FORCE ( F ) = ???

____ [ BY USING FORMULA ] ____

[ $F \: = \: i \: \times \: L \: \times \: B \: \times \: \sin( \alpha ) ]$

★ F = i × L × B × Sin 30°

» F = 5 × ( 5 × 10-² × 2.5 × 10-² )× 1 / 2

» F = 62.5 × 10-⁴ × 1 / 2

★ MAGNETIC FORCE ( F ) =

31.25 × 10-⁴ N ★

[ ✓ NOTE :- FORCE ON WIRE KEPT IN SOLENOID IS INDEPENDENT OF NO. OF TURNS OF COIL AND IT'S RADIUS ✓ ]

______________________________________

ANSWER :- MAGNETIC FORCE ON CURRENT CARRYING WIRE IN SOLENOID IS ;

★ F = 31.25 × 10-⁴ N ←

[ OR ]

★ F = 3.125 × 10-³ N ←

Answer 2

Given :

➟ L (length of the wire) = 5 cm = 5 × 10-² m

➟ Ø (angle) = 30°

➟ I (current in the wire) = 5 A

➟ B (Magnetic field) = 2.5 × 10-² T

Find :

Magnetic force on the wire (F).

Solution :

We know that

• F = I × B × L sinØ

Put the known values in above formula

→ F = 5 × 2.5 × 10-² × 5 × 10-² × sin30°

And sin30° = 1/2

→ F = 12.5 × 10-² × 5 × 10-² × 1/2

→ F = 62.5 × 10-² × 10-² × 2

→ F = 31.25 × 10-⁴

→ F = 3.125 × 10-³ N

∴ Magnetic force on the wire is 3.125 × 10-³ N.

(The given options are wrong)