JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Magnetic Effects of Current

A charge particle of

2 μ C

accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is ____

\times 1 0^{- 18} kg

Answer: 144

Solution

The speed gained across the potential difference follows from

q V = \frac{1}{2} m v^{2}

, and in the field the circular radius is

r = \frac{m v}{q B}

. Eliminating

v

gives

r = \frac{2 mV / q}{B}

, so

m = \frac{r ^{2} q B ^{2}}{2 V} .

With

r = 0.03 m

q = 2 \times 1 0^{- 6} C

B = 4 \times 1 0^{- 3} T

and

V = 100 V

m = \frac{( 0.03 ) ^{2} ( 2 \times 1 0 ^{- 6} ) ( 4 \times 1 0 ^{- 3} ) ^{2}}{2 ( 100 )} = 144 \times 1 0^{- 18} kg .

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Magnetic Effects of Current

Solution

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